While I have to admit it was quite tiring and time consuming, methodology wise, it was pretty simple. Not even comparable to some monster like lim( n!/n^n)^(1/n) with n approaching infinity. That’s by far the most difficult one I have ever seen
To be honest, if I got that in an exam, about half-way through working it out I'd assume that I'd done something wrong and just give up and move on to the next question.
Ben McKenzie This solution is unnecessary complicated, with all the fraction multiplying. Just collect x, do a MacLaurin expansion of the square roots, stopping at the first term after the 1 in order not to get 1^inf. It's going to be: lim (1 + a/x)^x = e^a, with a = -1/2, which you should know, or if you don't, you can easily find with De l'Hopital.
The teachers I have had in the past would hate that answer. They'd want to get the square roots out of the denominator, and have (square root e)/e instead.
This is an easy one. Next up, find the limit of the needed angles to perform 3 consecutive orthographic rotations in Polar Coordinates using Quaternions while mapping it to a 3D Complex Cartesian plane where those rotations do not change its original orientation!
@@kausarmeutuwah8304 Haha! I was being a bit sarcastic! That type of problem with all of the multivariable unknowns and many partial derivatives and multiple integrations would be insane! It would probably take 3-5 whiteboards and about 30 pens and about 15 hours to do it by hand! Let's just open up Wolfram or Mathlab... let it do it for us! Get the response, can not compute!
imagine you want to show this in a perfect formal way. I think the video would go over 1 hour. But a real good problem in which you can check if you really confident in limits.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
@@seroujghazarian6343 That's not hard to do. How many integers are there in the set of values in this inclusive range of [0,1]? Simply 2. Now, how many reals are there in that same inclusive range? Infinite!
@@seroujghazarian6343 Nice! I like my conjecture... Every single concept of mathematics, all branches, and levels are all embedded in the simple expression (1+1) ... Everything is derived or integrated from it. Just the act of adding 1 to itself, the application of applying the operation of addition which is a linear transformation, translation to be exact defines the unit circle. There is perfect symmetry, reflection, and a 180 degree or PI radians rotation embedded within it. It isn't directly obvious at first, but take a piece of paper and mark a point on it and draw a line segment of an arbitrary distance towards your right. Now label the starting point 0 and the ending point 1. To add 1 to this line segment or unit vector is to take the total length or its magnitude and translate it along the same line in the same direction. The tail of the new vector will be at the head of the original and the head of the new vector will be pointing at 2. By doing this the total distance is from the initial point of 0 to the new location will be 2. This turns the expression of (1+1) into the equation (1+1) = 2. This equation is actually the definition of both the Pythagorean Theorem and the Equation of the Unit Circle that is positioned at the origin (0,0). You see, we went to the right from the starting point and labeled that 1. We could of went to the left and labeled it 1 as well. However, they are opposing directions, and vectors have two parts, magnitude its length, and its sign or direction or angle of rotation. So, we can label the point to the left with -1. Here the starting point of 0 is the point of reflection, point of symmetry and the point of rotation. If we rotate the point 1 to the point -1 with respect to the initial point 0, you will make an arc that is PI radians which also takes you from 1D space into 2D space. Now we have the Y coordinates as well. This is simply due to 1+1 = 2 which is also 1x2 = 2. And this is evident because we know that a circle with a radius of one has a diameter of 2, its Circumference is 2*PI and its Area is PI units^2. We know that the Pythagorean Theorem is C^2 = A^2 + B^2. We also know that the equation of a circle centered at the origin is r^2 = x^2 + y^2... They are the same exact equation. When we look at the general equation of a line in the form of y = mx+b we know that m is the slope between two points and b is the y-intercept. The slope is m = (y2 - y1)/(x2-x1). We can let m = 1, and b = 0 and this gives us y = x. A diagonal line that goes through the origin. This line has an angle of 45 degrees or PI/4 radians above the X-axis. We know by definition that the slope is rise/run. We also see that it is (y2-y1)/(x2-x1) which is also dy/dx, change in y over change in x. If we look closer we can see that dy/dx is also sin(t)/cos(t) where (t) is the angle above the x-axis. This is also tan(t). When you look at the original equation of the line y = mx+b we can see this as f(x) = tan(t)x + b. This all comes from 1+1 = 2! Every polynomial, every geometrical shape, including vectors and matrices, their operations, even concepts in calculus such as derivatives and integrals are rooted in (1+1)! I just love how everything within mathematics is all connected and related! I'm sure you know all of these concepts but was just wanting to illustrate all of their interconnections. Yes, I take a Physical approach to mathematical induction! Why? Because without physics, or the ability to move, or translate, then the operation of addition would have no application or meaning! You can not even add 1 to itself in a scalar manner without treating them first as vector quantities. The number 1 itself is the unit vector, and the unit vector is the number 1 itself. You need physics, motion to perform the operation of addition! And as soon as you have motion, you have, limits, derivatives, and integrals!
This is honestly the best limit I have ever seen, it has literally everything a student needs to be able to calculate limits (when x goes to infinity). Absolutely amazing.
I just failed my calculus test, and after having watched so many videos on youtube on how to solve indeterminates, the algorithm has recommended me this. Never before have I been so happy to never have stumbled upon this equation. I'm pretty sure that it would have broken me if it had appeared in the test. Now I have newfound respect towards normal indeterminates. This right here is the eldritch equivalent of indeterminates. It makes me shudder to think how you even stumbled upon it.
As soon as I saw it, I knew it would lead to 1^infinity What I was not ready for was the bit after. I love your videos and they've genuinely helped me.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
@@pullingrabbitsouttaahat this link is probably an add to his content that seems controversial, probably not something to waste time one except if you wanna laugh.
By doing a little more algebra on the first limit calculation (that resulted in 1) you get 2x/(2x+1) (which is 1 - 1/(2x+1) ). So the original limit is (2x/(2x+1))^x. By substituting m=2x+1, this results in (1-1/m)^(m/2 - 1/2). Using lim as n->inf (1+a/n)^bn = e^ab, the limit is e^(-1/2).
You can do this problem with a Taylor approach, which ends up being formally written with oh notation (you need your error in the brackets to be o(1/x) to get the right final answer).
I saw this video on my recommended today and even though it is late I just had to comment on what I found. Changing the constants by 1 multiplies the solution by a factor of 1/√e As in lim(x->inf) of (√x²+2x+4-√x²+3)^x = 1 and lim(x->inf) of (√x²+2x+2-√x²+3)^x = 1/e The other constant also works similarly lim(x->inf) of (√x²+2x+3-√x²+2)^x = 1 and lim(x->inf) of (√x²+2x+3-√x²+4)^x = 1/e Basically the solution comes out as lim(x->inf) of (√x²+2x+a-√x²+b)^x = e^[½(a-b-1)] I would never have guessed that just by looking at the equation.
Not sure if this makes it any simpler but since you have f(x)^x as a limit, maybe you could use the limit that defines e^x? Or in this case, e^k? Like so... [sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3)]^x = [1 + k/x]^x solving for k gives k = x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1] but [1 + k/x]^x is just e^k in the limit you still have to work out that x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1] is equal to -1/2 in the limit
This gave me the idea of an inverse limit. I can't really come up with a way for it to be used consistently. "Lim^-1 as x -> oo of 2" would be the notation. Maybe, instead of a constant, it would help do a function inside a function?
I had that idea too, but then it quickly dawned upon me that the idea of an inverse limit is impossible. This is why:- 1) lim (2x/x) x->oo 2) lim (x²/x) x-> 2 3) lim (x) x->2 All lead to the same result, but there's no way that we can list all possible limits which lead to 2. You will say, we cannot list all the answers of ln(-1), but that's different. There's at least a system by which we can do this. We only have to keep changing the number of rotations. This case is exactly the result of 'there are different types of infinities'. You can say that ln(-1) has countably infinite answers, while the idea of inverse limit has uncountably many answers.
@@createyourownfuture3840 is it different to log1(x)? I do admit the inverse is realistically useless. My guess is if it DID have a use, it would more likely for organization (ie. Making sure that you have to raise your "answer" to e in order to fully answer the question. Maybe something with catagorization theory (I heard it's a thing, and I have no clue what it's about minus the obvious)?
I think I have a challenge for you... It's a question that was on the test for calculus monitor of my college, here in Brazil. It goes like this: -Calculate: limit when n -> Infinite of ( 1/((√n).(√(n+1))) + 1/((√n).(√(n+2))) + 1/((√n).(√(n+3))) + ... + 1/((√n).(√(n+n))) ). It might be a challenge, or not! It would be really nice to see you solving it in video, if possible! Thanks for all your videos, they are very funny and inspiring!
Yo Dawg, I heard you like indeterminate forms, so I put an indeterminate form on your indeterminate form so you can calculate limits while you calculate limits!
What a patient guy... I usually edit a step to change it into the next step and this way I save a lot of ink and space. But he is so hard working. Hats Off..
In India , we have a easy formula for limits in the form 1^Infinity. It is lim x->a f(x) ^g(x) and f(a)=1 and g(a) =Infinity = e^{lim x->a g(x) * ( 1 - f(x) )}
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
As someone who is not very familiar with calculus, I feel like bprp is making up rules as he goes to make it look like he knows what he's doing (which he is)
Another way, if you happen to know Taylor series, (1+x)^n = 1 + n*x + (1/2) * n(n-1)x^2 + ... n = 0.5 for square root, so sqrt(x^2 + 2x + 3) = x sqrt(1 + 2/x + 3/x^2) = x (1 + 0.5 (2/x+3/x^2) - (1/8) (2/x + 3/x^2)^2 + ... ) and sqrt(x^2 + 3) = x sqrt(1 + 3/x/x) = x (1 + 0.5 (3/x^2) - (1/8) (3/x^2)^2 + ... ) As x -> infinity, 1/x >> 1/x^2 >> 1/x^3, so we can ignore higher terms once we have a few non-zero terms. Subtracting the two, term by term: x (1 + 0.5 (2/x+3/x^2) - (1/8) (2/x + 3/x^2)^2 + ... ) - x (1 + 0.5 (3/x^2) - (1/8) (3/x^2)^2 + ... ) =x (0 + 0.5(2/x) - (1/8) (4/x^2 + 12/x^3 + 9/x^4 - 9/x^4)) =x ( 1/x - (1/2x^2) - (1/x^3 terms and smaller, which can be ignored)) = 1 - (1/2)(1/x) - (smaller terms, which go to 0, taking x -> infinity) The limit is then quickly (1 - (1/2)(1/x))^x, which we know to be e^-(1/2). Solving it this way means thinking of a few less-tedious steps. 1. A few terms of Taylor series, plugging into a general formula we already know 2. Subtract the two parts to see what is left over. 3. Make sure what is left over is not zero. Add more terms and repeat step 1 if it is zero. In this case, I expanded to 1/x^2, ignoring 1/x^3., and the components of the power series were (2/x + 3/x^2)^n, so I knew I needed Taylor series terms up to n = 2. If I used only n = 1, I would have missed the contribution (2/x + 3/x^2)^2 = (4/x^2 + ...). If I didn't expand to 1/x^2, I would have been left with (1)^infinity, which is indeterminate. I knew this would happen because I had something that looked like x sqrt (1 + ... ) - x sqrt (1 + ... ). 4. Use definition of e. The time-consuming part is repeatedly adding more terms to the Taylor expansion. Intuition and practice will help you figure out how many terms and what order to expand to. Most of the time the answer will only require n = 1 or n = 2, requiring multiplications instead of derivatives, and usually requiring only 1/x or 1/x^2 terms. The point is: Taylor series can do the derivatives for you, so all you do is multiply terms and keep track of the ones that are large enough to matter.
Doing it this way helps to see what would happen if the numbers were replaced with other numbers. lim ( sqrt(x^2 + 2x + a) - sqrt(x^2 + b) )^x = e^((a-b-1)/2) for any real a,b. The 2x needs the 2 because of the square root. Hope you make a cube root (x^3) version next >:)
Omg!! I've done it. The answer is 1. I just aplied LH two times and i ended with ln(L)=0 so L is equal to 1. I take a look with geogebra and im right! I can't believe this is the first limit i've solved
Could you take the limit when x goes to negative infinty and work with complex numbers? I think that would be pretty dificult (Sorry for the bad english)
I don't mind answering this question in my final exam but only if you are my calculus teacher, for I would then surely know that my efforts would be truly appreciated. Of course no other calculus teacher would even dream of such a vile limit, let alone put it on a final exam paper. Great explanation, great problem and as usual - that's it.
At 5:30, when using the x^2 pieces of sqrt to cancel the 2x, how can you ignore the remaining x^1 term, that would give some factor of sqrt(inf) in the denominator, which would send the limit to zero?
when solving limits where x-->inf , especially with functions that are one polynomial over another, a useful trick is to multiply both the numerator and the denominator by (1/x). keep doing this until the highest degree term is reduced to only a coefficient. then, when the limit is taken, those other terms end up being some number devided by infinity, which makes them zero. this is why he ignores all but the highest degree term; the other terms are reduced to zero.
@@badhbhchadh most likely he didn't use a ti89, i typed it in like 2 minutes ago and it's still calculating LMAO wolfram alpha gives it in 2 seconds tho
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
At 6:20 roughly I get confused; shouldn’t the limit be to 0 since we know both sqrt(x^2 +2x + 3) and sqrt(x^2 + 3) are both >x so the denominator is > 2x and the fraction is
Nicholas 4321 they are equivalent for large x values. more formally 2x/(sqrt(x^2+2x+3) +sqrt(x^2+3) = 2x/(x(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)))= 2/(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)) as x goes to infinity 2/x and 3/x^2 go to 0 (if this seems non-obvious, think about what happens when you divide something by a really big number, or look at their graphs) so we get 2/(sqrt(1)+sqrt(1))=2/2= 1
here's a simple explanation from a 10th grader: (personal opinion - theory) * infinity is not a number, therefore something unknown, probably an unknown set with unknown value. it's a value that's above every number. therefore: 1. subtracting infinity from infinity equals infinity and not 0 again, due to the fact that its value is unknown. upon raising it to the power of infinity, we would again get infinity. 2.another theory is that subtracting infinity from infinity may equal an infinity smaller than those 2. and that smaller infinity can eventually get into a really big one if we raise it the power of infinity. (ok this one's a little confusing lol) 3. finally, if infinity - infinity can equal 0, raising it to the power of infinity can somehow equal infinity again, similar with the undefined equation 0x = a (a not 0). 4. considering that infinity may sometimes represent the set of real numbers, we can use simple algebra (which is false) to get what we know from properties: (infinity - infinity)^infinity = 0^infinity = 0 5. another theory may be that again if infinity represents the real number set, we can say that we have a subtraction within all 2 real numbers raised to a power that has all the real numbers. that sounds like the 2nd one, except we have a specific number set, the real one. remember those are theories, nothing of them is proved so they're false in current terms of simple algebra
At the step where we get 2x/2x, can we not just conclude that the bottom part is strictly larger than 2x (since we ignored some parts and the square rote is strictly monotone increasing) and thus conclude that we are approaching 1 from the bottom, and since we are strictly smaller than 1, the limit of that to X is 0?
how do you know only to take x^2 terms from the denominator at 5:45? i thought he was applying L'hopital's rule but i checked, and you cant solve it by that because the differentiation keeps getting more and more complicated
If I was a calculus professor I would put this limit in as a bonus question for a test for extra credit. It’s a good challenge for the students who could take it on.
at 13:08 where does the x come from? didn’t you cancel out the two to get one in the numerator before? shouldn’t you get -sqrt(x^2+2x+3) rather than that with an x in front?
I liked the video, thanks! What allows us to ignore the '2x' part of the sqrt(x^2 + 2x + 3) when looking at the limits? I get ignoring the + 3's because they really don't matter, but the 2x part??? I think you were right to do it but it just would be better if we knew when it is safe to use the shortcuts you used and when it will get us in trouble.
@@lossen1984 Imagine x is 9999 or some really big number. 9999^2 is 9999 times bigger than 9999. So when you eventually hit infinity, the x^2 is infinitely bigger than x, so you can ignore x.
I don't seem to agree with something proposed in 1:06. Because he analyses both limits separately, as if they were convergent. Of course if you have 2^limx->inf {x} you get infinity, but if u have some function f(x)^x, when x->inf, you can't just say because limx->inf{f(x)}=2 then limx->inf {f(x)^x} = 2^limx->inf{x}. Because to do so, both limits must exist. My point is, may be limx->inf {f(x)^x} exists, even if limx->inf{f(x)} isn't one, or doesn't exist. Am i right? By the way, i love your videos, please keep teaching us.
An easier way to solve this would be to recognize that x^2 +2x +3 = (x+1)^2 + 2. Then when you calculate the inner limit, keep an extra term. [(x^2 +2x +3) - (x^2 +3)]÷[((x+1)^2 +2)^.5 + (x^2 +3)^.5] -> (2x) ÷ (x+1 + x) = 1 - 1/(2x +1). When you then calculate this to the power x and compare it to the definition of e, the answer e^(-1/2) drops out automatically. To be more careful, you can show that the inner function is equal to (1 - 1/(2x) + O(1/x^2)) for all positive x.
I had an idea of inf-inf So I tired it with ln( ) - ln( ) but it was kind weird so I tried with sqrt( ) - sqrt( ) and i wanted it to end up to be 1. Then raise that to the x power. Done : )
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
Or could have just substituted x=1/t Then used binomial on those two polynomials. Finally, series expansion of ln(1-t) or standard formula of ln(1+x)/x = 1. Could have been much much easier to solve but ok
i have a rule made its called de' or eo rule (de)means take the derivative of all functions inside the parentheses and the derivative of the power as well, eo means take the integral ( the second step might not work, but you should try ot if the de step fails )
how to solve this within 2mins speedrun : first take the limit of the base, you find it to be 1, so its 1^inf form lim f^g when f --> 1 and g--> inf is e^lim (f-1)*g use this standard trick, and you get a simple limit of 0 * inf form use binomial thereom for general powers find the constant term and ignore terms with lower powers of x, you get -1/2 Enjoy your correct answer.
Why are you not using a second time the "check" in the 4th line? Cause at the limit (x+1)(x^2+3)^(1/2)-x (x^2+2x+3)^( 1/2)=-x+ (x^2+3)^(1/2) now i guess it is easier to know this limit
Because you are supposed to keep doing all the algebra & arithmetic that you can legally do before taking the limit. For example, if you have x^3/x^2 you will do some cancellation before taking the limit correct? So this would just be x but if you took the limit, say to 0, then this would be 0/0 which is indeterminate. But since we have some legal operations available, namely we have three x's on top & can get rid of the two sitting at the bottom, it leaves us with just one x. I know its a simple ass example but just trying to show the concept.
Why does calculus make it so complicated, infinity mean it never run out, so if you treated infinity like a number infinity -infinity would just equal infinity, the same concept that infinity can never go higher then infinity, so if you add infinity to infinity it is still infinity, the same if you have infinity multiplied by infinity, still just equal infinity, the same think if you say that infinity multiply by 0 equal 0, if infinity is treated like a number then you are wrong because your breaking the law of infinity, infinity mean it never run out so it will never run out even if multiplies, calculus break the main rules of infinity it cause infinity to run out by becoming an imaginary number. If infinity is not treated as a number and infinity multiplies by 0 with no calculus non-sense it would equal 0, but that would be the same as x multiply by 0 equal 0, the same this way if not treated as a number and also no calculus, so infinity -infinity =0 so it is just treated the same as if it was x-x=0
Sorry for my unjustified cockyness :-/ Of course there’s supposed to be -1 there. Need to follow the thought process more carefully and don’t question the expert’s competence... Keep it up maestro :-D
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
The limit of 1/x as x→∞ is Zero ... but 1/∞ is not zero, the limit of it is Zero. Meaning 1/∞ never gets to Zero, just arbitrarily close. 1/∞ is Asymptotically Equivalent to Zero. Asymptotic means approaching a value or curve arbitrarily closely (i.e., as some sort of limit is taken).
could you expand the 2 square roots using the binomial theorem, get some cancellations when taking the difference, and somehow put a cap on the remainder?
the square root on the left grows faster because there's an extra +2x term. But I'm not sure how you could conclude the entire thing is equal to one, just from that.
This is the vilest limit I've ever seen and if I ever become a calculus teacher I'll definitely put it in one of the exams
Kiritsu u evil beast
Just make it the exam. Or the extra credit that gives you a huge boost.
While I have to admit it was quite tiring and time consuming, methodology wise, it was pretty simple. Not even comparable to some monster like lim( n!/n^n)^(1/n) with n approaching infinity. That’s by far the most difficult one I have ever seen
@@giannispolychronopoulos2680 yea that one is definition of hard question-simple answer.
not good not good we need more proofs related questions
To be honest, if I got that in an exam, about half-way through working it out I'd assume that I'd done something wrong and just give up and move on to the next question.
I mean the exam would have to be just that question because you're using every rule in it anyway haha
Ben McKenzie This solution is unnecessary complicated, with all the fraction multiplying.
Just collect x, do a MacLaurin expansion of the square roots, stopping at the first term after the 1 in order not to get 1^inf. It's going to be:
lim (1 + a/x)^x = e^a, with a = -1/2,
which you should know, or if you don't, you can easily find with De l'Hopital.
landochabod7 I have no idea what you’re talking about, and this is from somebody who understands everything bprp’s talking about!
I thought the same.
@@tcocaine it would be best to give this as a homework exercise.
Bursts out laughing when you said pray your calculus 2 teacher doesn’t see this. Thank goodness I already finished calc 2
nadia salem
(I am actually a calc2 teacher loll)
blackpenredpen I just hope you don’t give this one on the finale unless it’s the only question
It's been 6 years since I took calc 2, although my teachers covered this type of limit in calc 1.
The teachers I have had in the past would hate that answer. They'd want to get the square roots out of the denominator, and have (square root e)/e instead.
Perhaps. In Calculus, it's customary to leave the result unrationalized.
after doing a question like that i wouldnt give a damn about rationalising the fraction lol
You mean e^1/2 ?
@@Nonexistility They just rationalized it
That's the difference between an algebra teacher and a calculus teacher.
If you look up the word "tedious" in the dictionary, you will almost certainly find a picture of this limit problem.
John Dixon
If you search calc2 final exam, then... :)
This is an easy one. Next up, find the limit of the needed angles to perform 3 consecutive orthographic rotations in Polar Coordinates using Quaternions while mapping it to a 3D Complex Cartesian plane where those rotations do not change its original orientation!
@@kausarmeutuwah8304 Haha! I was being a bit sarcastic! That type of problem with all of the multivariable unknowns and many partial derivatives and multiple integrations would be insane! It would probably take 3-5 whiteboards and about 30 pens and about 15 hours to do it by hand! Let's just open up Wolfram or Mathlab... let it do it for us! Get the response, can not compute!
Heavyyyyyy!!!!
Heavyyyyyy!!!!
imagine you want to show this in a perfect formal way. I think the video would go over 1 hour. But a real good problem in which you can check if you really confident in limits.
namensindüberbewertet yea I know. That's why I just circled circled loll
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
@@pullingrabbitsouttaahat I would see it but you spelled l'Hôpital wrong
@@JakubS Thanks For Pointing Out. But I Don't Care Much Useless Things.
Square root of e... That's when you know you broke mathematics.
Or when you find out that a subset has more elements than the original set
@@seroujghazarian6343 That's not hard to do. How many integers are there in the set of values in this inclusive range of [0,1]? Simply 2. Now, how many reals are there in that same inclusive range? Infinite!
@@skilz8098 Yeah, but considering ]0,1[ has as many elements as R (cot(πx) being the bijection between them)...
@@seroujghazarian6343 Nice!
I like my conjecture...
Every single concept of mathematics, all branches, and levels are all embedded in the simple expression (1+1) ...
Everything is derived or integrated from it. Just the act of adding 1 to itself, the application of applying the operation of addition which is a linear transformation, translation to be exact defines the unit circle. There is perfect symmetry, reflection, and a 180 degree or PI radians rotation embedded within it.
It isn't directly obvious at first, but take a piece of paper and mark a point on it and draw a line segment of an arbitrary distance towards your right. Now label the starting point 0 and the ending point 1. To add 1 to this line segment or unit vector is to take the total length or its magnitude and translate it along the same line in the same direction. The tail of the new vector will be at the head of the original and the head of the new vector will be pointing at 2.
By doing this the total distance is from the initial point of 0 to the new location will be 2. This turns the expression of (1+1) into the equation (1+1) = 2. This equation is actually the definition of both the Pythagorean Theorem and the Equation of the Unit Circle that is positioned at the origin (0,0).
You see, we went to the right from the starting point and labeled that 1. We could of went to the left and labeled it 1 as well. However, they are opposing directions, and vectors have two parts, magnitude its length, and its sign or direction or angle of rotation. So, we can label the point to the left with -1.
Here the starting point of 0 is the point of reflection, point of symmetry and the point of rotation. If we rotate the point 1 to the point -1 with respect to the initial point 0, you will make an arc that is PI radians which also takes you from 1D space into 2D space. Now we have the Y coordinates as well.
This is simply due to 1+1 = 2 which is also 1x2 = 2. And this is evident because we know that a circle with a radius of one has a diameter of 2, its Circumference is 2*PI and its Area is PI units^2.
We know that the Pythagorean Theorem is C^2 = A^2 + B^2. We also know that the equation of a circle centered at the origin is r^2 = x^2 + y^2... They are the same exact equation.
When we look at the general equation of a line in the form of y = mx+b we know that m is the slope between two points and b is the y-intercept. The slope is m = (y2 - y1)/(x2-x1).
We can let m = 1, and b = 0 and this gives us y = x. A diagonal line that goes through the origin. This line has an angle of 45 degrees or PI/4 radians above the X-axis.
We know by definition that the slope is rise/run. We also see that it is (y2-y1)/(x2-x1) which is also dy/dx, change in y over change in x. If we look closer we can see that dy/dx is also sin(t)/cos(t) where (t) is the angle above the x-axis. This is also tan(t).
When you look at the original equation of the line y = mx+b we can see this as f(x) = tan(t)x + b. This all comes from 1+1 = 2!
Every polynomial, every geometrical shape, including vectors and matrices, their operations, even concepts in calculus such as derivatives and integrals are rooted in (1+1)!
I just love how everything within mathematics is all connected and related! I'm sure you know all of these concepts but was just wanting to illustrate all of their interconnections.
Yes, I take a Physical approach to mathematical induction! Why? Because without physics, or the ability to move, or translate, then the operation of addition would have no application or meaning! You can not even add 1 to itself in a scalar manner without treating them first as vector quantities. The number 1 itself is the unit vector, and the unit vector is the number 1 itself. You need physics, motion to perform the operation of addition! And as soon as you have motion, you have, limits, derivatives, and integrals!
skilz8098 very well written!
This is honestly the best limit I have ever seen, it has literally everything a student needs to be able to calculate limits (when x goes to infinity). Absolutely amazing.
I just failed my calculus test, and after having watched so many videos on youtube on how to solve indeterminates, the algorithm has recommended me this.
Never before have I been so happy to never have stumbled upon this equation. I'm pretty sure that it would have broken me if it had appeared in the test. Now I have newfound respect towards normal indeterminates. This right here is the eldritch equivalent of indeterminates. It makes me shudder to think how you even stumbled upon it.
As soon as I saw it, I knew it would lead to 1^infinity
What I was not ready for was the bit after. I love your videos and they've genuinely helped me.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
@@pullingrabbitsouttaahat this link is probably an add to his content that seems controversial, probably not something to waste time one except if you wanna laugh.
This is truly great maths to watch, and one hell of a limit problem. Thanks for posting.
Lies again? DMP Triple
By doing a little more algebra on the first limit calculation (that resulted in 1) you get 2x/(2x+1) (which is 1 - 1/(2x+1) ). So the original limit is (2x/(2x+1))^x. By substituting m=2x+1, this results in (1-1/m)^(m/2 - 1/2). Using lim as n->inf (1+a/n)^bn = e^ab, the limit is e^(-1/2).
This is the best answer
The evil Laugh at 17:18 x)
: )
This is those questions that you skip without even looking at it
I just finished my calc I final exam and I gotta say I’m pretty excited for calc II. Thank you for such a great rigorous video to end my night :D
congrats on finishing calc 1
This idea of ignoring lower order terms reminds me of something we call O-notation in rough algorithmic time and memory measurement calculations.
You can do this problem with a Taylor approach, which ends up being formally written with oh notation (you need your error in the brackets to be o(1/x) to get the right final answer).
Thanks!
I am really happy to see 20 minutes video. Yay indeed! Thanks!
Crazy Drummer I am very glad too. Thank you!!!
Maths will be interesting if you are my teacher. I remembered that today is teachers day.
Happy teachers day
I saw this video on my recommended today and even though it is late I just had to comment on what I found.
Changing the constants by 1 multiplies the solution by a factor of 1/√e
As in lim(x->inf) of (√x²+2x+4-√x²+3)^x = 1 and lim(x->inf) of (√x²+2x+2-√x²+3)^x = 1/e
The other constant also works similarly lim(x->inf) of (√x²+2x+3-√x²+2)^x = 1 and lim(x->inf) of (√x²+2x+3-√x²+4)^x = 1/e
Basically the solution comes out as lim(x->inf) of (√x²+2x+a-√x²+b)^x = e^[½(a-b-1)]
I would never have guessed that just by looking at the equation.
Am I the only one that got cracked up when he said "oh, I drew the same box again" I spent 5 minutes plus laughing man
Not sure if this makes it any simpler but since you have f(x)^x as a limit, maybe you could use the limit that defines e^x? Or in this case, e^k? Like so...
[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3)]^x = [1 + k/x]^x
solving for k gives
k = x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1]
but [1 + k/x]^x is just e^k in the limit
you still have to work out that x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1] is equal to -1/2 in the limit
This gave me the idea of an inverse limit. I can't really come up with a way for it to be used consistently. "Lim^-1 as x -> oo of 2" would be the notation. Maybe, instead of a constant, it would help do a function inside a function?
Do you mean "find a function that approaches this limit as x->inf"?
@@PixelSergey I'm not really sure.
I had that idea too, but then it quickly dawned upon me that the idea of an inverse limit is impossible. This is why:-
1) lim (2x/x)
x->oo
2) lim (x²/x)
x-> 2
3) lim (x)
x->2
All lead to the same result, but there's no way that we can list all possible limits which lead to 2. You will say, we cannot list all the answers of ln(-1), but that's different. There's at least a system by which we can do this. We only have to keep changing the number of rotations. This case is exactly the result of 'there are different types of infinities'. You can say that ln(-1) has countably infinite answers, while the idea of inverse limit has uncountably many answers.
@@createyourownfuture3840 is it different to log1(x)? I do admit the inverse is realistically useless. My guess is if it DID have a use, it would more likely for organization (ie. Making sure that you have to raise your "answer" to e in order to fully answer the question. Maybe something with catagorization theory (I heard it's a thing, and I have no clue what it's about minus the obvious)?
If we substitute x with 1/y , y approaching 0 in the positive area we can jump a lot of algebric steps.
This 22 mins was the best part of my day...
I think I have a challenge for you...
It's a question that was on the test for calculus monitor of my college, here in Brazil. It goes like this:
-Calculate: limit when n -> Infinite of ( 1/((√n).(√(n+1))) + 1/((√n).(√(n+2))) + 1/((√n).(√(n+3))) + ... + 1/((√n).(√(n+n))) ).
It might be a challenge, or not! It would be really nice to see you solving it in video, if possible!
Thanks for all your videos, they are very funny and inspiring!
Yo Dawg, I heard you like indeterminate forms, so I put an indeterminate form on your indeterminate form so you can calculate limits while you calculate limits!
I just want to know who drew the house that you can faintly see on the left side of the whiteboard
What a patient guy...
I usually edit a step to change it into the next step and this way I save a lot of ink and space. But he is so hard working. Hats Off..
In India , we have a easy formula for limits in the form 1^Infinity.
It is lim x->a f(x) ^g(x) and f(a)=1 and g(a) =Infinity
= e^{lim x->a g(x) * ( 1 - f(x) )}
Yaa its right
that was süper brütal
also, as a christmas gift almost as good as socks.
AndDiracisHisProphet hahahahahah. I like those dots on the u
that's the second time derivative.
Wats wrong with socks? I could use a few pairs. :)
It's like gifting your girlfriend with a vacuum cleaner or an iron
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
I'm glad that you're not a lecturer in my univ, or else the exam will be hellish haha
Max Haibara tbh I think this problem would be fun but it would take me hours at least
As someone who is not very familiar with calculus, I feel like bprp is making up rules as he goes to make it look like he knows what he's doing (which he is)
Watched the video twice (and know calc) everything he does is correct
I just would have done it differently
Another way, if you happen to know Taylor series, (1+x)^n = 1 + n*x + (1/2) * n(n-1)x^2 + ...
n = 0.5 for square root, so
sqrt(x^2 + 2x + 3) = x sqrt(1 + 2/x + 3/x^2) = x (1 + 0.5 (2/x+3/x^2) - (1/8) (2/x + 3/x^2)^2 + ... )
and sqrt(x^2 + 3) = x sqrt(1 + 3/x/x) = x (1 + 0.5 (3/x^2) - (1/8) (3/x^2)^2 + ... )
As x -> infinity, 1/x >> 1/x^2 >> 1/x^3, so we can ignore higher terms once we have a few non-zero terms.
Subtracting the two, term by term:
x (1 + 0.5 (2/x+3/x^2) - (1/8) (2/x + 3/x^2)^2 + ... )
- x (1 + 0.5 (3/x^2) - (1/8) (3/x^2)^2 + ... )
=x (0 + 0.5(2/x) - (1/8) (4/x^2 + 12/x^3 + 9/x^4 - 9/x^4))
=x ( 1/x - (1/2x^2) - (1/x^3 terms and smaller, which can be ignored))
= 1 - (1/2)(1/x) - (smaller terms, which go to 0, taking x -> infinity)
The limit is then quickly (1 - (1/2)(1/x))^x, which we know to be e^-(1/2).
Solving it this way means thinking of a few less-tedious steps.
1. A few terms of Taylor series, plugging into a general formula we already know
2. Subtract the two parts to see what is left over.
3. Make sure what is left over is not zero. Add more terms and repeat step 1 if it is zero.
In this case, I expanded to 1/x^2, ignoring 1/x^3., and the components of the power series were (2/x + 3/x^2)^n, so I knew I needed Taylor series terms up to n = 2. If I used only n = 1, I would have missed the contribution (2/x + 3/x^2)^2 = (4/x^2 + ...). If I didn't expand to 1/x^2, I would have been left with (1)^infinity, which is indeterminate. I knew this would happen because I had something that looked like x sqrt (1 + ... ) - x sqrt (1 + ... ).
4. Use definition of e.
The time-consuming part is repeatedly adding more terms to the Taylor expansion. Intuition and practice will help you figure out how many terms and what order to expand to. Most of the time the answer will only require n = 1 or n = 2, requiring multiplications instead of derivatives, and usually requiring only 1/x or 1/x^2 terms.
The point is: Taylor series can do the derivatives for you, so all you do is multiply terms and keep track of the ones that are large enough to matter.
Doing it this way helps to see what would happen if the numbers were replaced with other numbers.
lim ( sqrt(x^2 + 2x + a) - sqrt(x^2 + b) )^x = e^((a-b-1)/2) for any real a,b. The 2x needs the 2 because of the square root. Hope you make a cube root (x^3) version next >:)
This is more of a hand workout than a calculus question 😂😂😂
Whats the limit of this expresion when x goes to 0? It is a 0^0 situation
omg.....
Omg!! I've done it.
The answer is 1.
I just aplied LH two times and i ended with ln(L)=0 so L is equal to 1. I take a look with geogebra and im right! I can't believe this is the first limit i've solved
OH WOW! that's impressive!!!
Could you take the limit when x goes to negative infinty and work with complex numbers? I think that would be pretty dificult
(Sorry for the bad english)
@@gregorio8827 I, using my s___ty math ended up with 0, read my comment to see how I got it
I don't mind answering this question in my final exam but only if you are my calculus teacher, for I would then surely know that my efforts would be truly appreciated. Of course no other calculus teacher would even dream of such a vile limit, let alone put it on a final exam paper. Great explanation, great problem and as usual - that's it.
At 5:30, when using the x^2 pieces of sqrt to cancel the 2x, how can you ignore the remaining x^1 term, that would give some factor of sqrt(inf) in the denominator, which would send the limit to zero?
when solving limits where x-->inf , especially with functions that are one polynomial over another, a useful trick is to multiply both the numerator and the denominator by (1/x). keep doing this until the highest degree term is reduced to only a coefficient. then, when the limit is taken, those other terms end up being some number devided by infinity, which makes them zero. this is why he ignores all but the highest degree term; the other terms are reduced to zero.
Yo i checked in the calculator and the limit is true!!
Itamar Rosen thanks!!!!!!
Damn, do you have a TI-89?
@@badhbhchadh most likely he didn't use a ti89, i typed it in like 2 minutes ago and it's still calculating LMAO
wolfram alpha gives it in 2 seconds tho
It means that you don't trust him.
Rationalizing denominator for the final Answer?
ZAID SALAMEH why?
How?
scrt(e)/e still has an irrational denominator
ZAID SALAMEH "we're adults now"
jeromesnail (sqrt(e))^(-1)
9:39 "because of the Chain Rule"
... the what?
Is that related to the Chen Lu?
No because of the china flu
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
lim x->infinity (1+1/x)^x->e; after we have 1^(infinity)=e
e comes up when you do not expect it. Brilliant!
lim (sqrt ax^2+bx+c-sqrt ax^2+c=e^-(b/4sqrt of a)
I've watched so many blackpenredpen videos enough to start understanding what it is I'm watching, I am so happy ~
At 6:20 roughly I get confused; shouldn’t the limit be to 0 since we know both sqrt(x^2 +2x + 3) and sqrt(x^2 + 3) are both >x so the denominator is > 2x and the fraction is
Nicholas 4321 they are equivalent for large x values. more formally 2x/(sqrt(x^2+2x+3) +sqrt(x^2+3) = 2x/(x(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)))= 2/(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)) as x goes to infinity 2/x and 3/x^2 go to 0 (if this seems non-obvious, think about what happens when you divide something by a really big number, or look at their graphs) so we get 2/(sqrt(1)+sqrt(1))=2/2= 1
Nicholas Jenkins gotcha eliminating x from either the numerator or denominator seems to make it a lot easier thanks 😊
here's a simple explanation from a 10th grader: (personal opinion - theory)
* infinity is not a number, therefore something unknown, probably an unknown set with unknown value. it's a value that's above every number. therefore:
1. subtracting infinity from infinity equals infinity and not 0 again, due to the fact that its value is unknown. upon raising it to the power of infinity, we would again get infinity.
2.another theory is that subtracting infinity from infinity may equal an infinity smaller than those 2. and that smaller infinity can eventually get into a really big one if we raise it the power of infinity. (ok this one's a little confusing lol)
3. finally, if infinity - infinity can equal 0, raising it to the power of infinity can somehow equal infinity again, similar with the undefined equation 0x = a (a not 0).
4. considering that infinity may sometimes represent the set of real numbers, we can use simple algebra (which is false) to get what we know from properties:
(infinity - infinity)^infinity = 0^infinity = 0
5. another theory may be that again if infinity represents the real number set, we can say that we have a subtraction within all 2 real numbers raised to a power that has all the real numbers. that sounds like the 2nd one, except we have a specific number set, the real one.
remember those are theories, nothing of them is proved so they're false in current terms of simple algebra
upside-down thumbnail
At the step where we get 2x/2x, can we not just conclude that the bottom part is strictly larger than 2x (since we ignored some parts and the square rote is strictly monotone increasing) and thus conclude that we are approaching 1 from the bottom, and since we are strictly smaller than 1, the limit of that to X is 0?
I worked it out with the simple approximation (a+b)^.5 is approximately a^.5+b/(2a^.5) when a>>b.
Best Xmas gift ever!
how do you know only to take x^2 terms from the denominator at 5:45?
i thought he was applying L'hopital's rule but i checked, and you cant solve it by that because the differentiation keeps getting more and more complicated
Thank god I finished all calculus courses before blackandredpen invented this monster
If I was a calculus professor I would put this limit in as a bonus question for a test for extra credit. It’s a good challenge for the students who could take it on.
at 13:08 where does the x come from? didn’t you cancel out the two to get one in the numerator before? shouldn’t you get -sqrt(x^2+2x+3) rather than that with an x in front?
He only canceled the 2s on top and bottom of both fractions. The x is still on the top.
Oh... my... God. That's some seriously insane algebra. Talk about a test of knowledge of detail! Well done.
I liked the video, thanks! What allows us to ignore the '2x' part of the sqrt(x^2 + 2x + 3) when looking at the limits? I get ignoring the + 3's because they really don't matter, but the 2x part??? I think you were right to do it but it just would be better if we knew when it is safe to use the shortcuts you used and when it will get us in trouble.
Yea why???
@@lossen1984 Imagine x is 9999 or some really big number. 9999^2 is 9999 times bigger than 9999. So when you eventually hit infinity, the x^2 is infinitely bigger than x, so you can ignore x.
@@lunarscapes6016 wow is it that simple? Well thanks for the explanation! :))
I ll have calculus I final exam next week. Thank you for the video!
Not quite understand of the video of time 5:10 so which video can explain this?Also not quite understand why no need absolute value?
I don't seem to agree with something proposed in 1:06. Because he analyses both limits separately, as if they were convergent. Of course if you have 2^limx->inf {x} you get infinity, but if u have some function f(x)^x, when x->inf, you can't just say because limx->inf{f(x)}=2 then limx->inf {f(x)^x} = 2^limx->inf{x}. Because to do so, both limits must exist. My point is, may be limx->inf {f(x)^x} exists, even if limx->inf{f(x)} isn't one, or doesn't exist. Am i right?
By the way, i love your videos, please keep teaching us.
Somehow this video makes me feel like I just bought a used car from a guy who talks very fast.
one of the most insane limits ever seen! a huge madness..
i have no idea why i'm watching this, but it's so interesting i stayed the entire 22 minutes
My Alma Mater, UC Berkeley. Love it, hate it. It's academics are great, love your lectures. Hate Berkeley's politics. I graduated deplorable. EE 1984.
I believe it would be easier to solve using t=1/x substitution and then doing McLauren series expansion at point t=0
An easier way to solve this would be to recognize that x^2 +2x +3 = (x+1)^2 + 2. Then when you calculate the inner limit, keep an extra term. [(x^2 +2x +3) - (x^2 +3)]÷[((x+1)^2 +2)^.5 + (x^2 +3)^.5] -> (2x) ÷ (x+1 + x) = 1 - 1/(2x +1). When you then calculate this to the power x and compare it to the definition of e, the answer e^(-1/2) drops out automatically. To be more careful, you can show that the inner function is equal to (1 - 1/(2x) + O(1/x^2)) for all positive x.
How did you come up with the original problem?
I had an idea of inf-inf
So I tired it with ln( ) - ln( ) but it was kind weird so I tried with sqrt( ) - sqrt( ) and i wanted it to end up to be 1.
Then raise that to the x power. Done
: )
did you expect the answer to be that beautiful or did you plan that?
You're a genius.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
I watch it all
amazing and very complicated differentiation work
thank you
Sqrt(e)/e makes my calc teacher happier. Great work amazing video!
Or could have just substituted x=1/t
Then used binomial on those two polynomials. Finally, series expansion of ln(1-t) or standard formula of ln(1+x)/x = 1. Could have been much much easier to solve but ok
Beautiful proof! Great job
4:58 Which other video is he talking about?
omg dude you are brUral with your algebra
I love the way he says square root
Next 1+1 = log 5 (69)/ln 2 e^x dx - lim 3->x^2
is a Indeterminaception XD
A great feat of perseverance using various interesting techniques. Awesome 👌
13:05 the x in the line above is hidden by the glare of the lights in case you're confused like I was
The limit of inside without any calculations should be 1, you can use Murphy's Law for it.
Hahahahah
This thing looks like the raidboss of calculus, but once we know its secret, it becomes easy(er)...... it still looks terrifying :D
i have a rule made its called de' or eo rule (de)means take the derivative of all functions inside the parentheses and the derivative of the power as well, eo means take the integral ( the second step might not work, but you should try ot if the de step fails )
What a brilliant sum ❤
how to solve this within 2mins speedrun :
first take the limit of the base, you find it to be 1, so its 1^inf form
lim f^g when f --> 1 and g--> inf is e^lim (f-1)*g
use this standard trick, and you get a simple limit of 0 * inf form
use binomial thereom for general powers find the constant term and ignore terms with lower powers of x, you get -1/2
Enjoy your correct answer.
Why are you not using a second time the "check" in the 4th line? Cause at the limit (x+1)(x^2+3)^(1/2)-x (x^2+2x+3)^( 1/2)=-x+ (x^2+3)^(1/2) now i guess it is easier to know this limit
At 14:34, wouldn't those x terms be infinity over infinity? Why is it valid to cancel them out?
Because you are supposed to keep doing all the algebra & arithmetic that you can legally do before taking the limit. For example, if you have x^3/x^2 you will do some cancellation before taking the limit correct? So this would just be x but if you took the limit, say to 0, then this would be 0/0 which is indeterminate. But since we have some legal operations available, namely we have three x's on top & can get rid of the two sitting at the bottom, it leaves us with just one x. I know its a simple ass example but just trying to show the concept.
Daychikho
Un=n^n
Tongs=u1+.................................... .. .............+undangtonquat
Why does calculus make it so complicated, infinity mean it never run out, so if you treated infinity like a number infinity -infinity would just equal infinity, the same concept that infinity can never go higher then infinity, so if you add infinity to infinity it is still infinity, the same if you have infinity multiplied by infinity, still just equal infinity, the same think if you say that infinity multiply by 0 equal 0, if infinity is treated like a number then you are wrong because your breaking the law of infinity, infinity mean it never run out so it will never run out even if multiplies, calculus break the main rules of infinity it cause infinity to run out by becoming an imaginary number. If infinity is not treated as a number and infinity multiplies by 0 with no calculus non-sense it would equal 0, but that would be the same as x multiply by 0 equal 0, the same this way if not treated as a number and also no calculus, so infinity -infinity =0 so it is just treated the same as if it was x-x=0
Do I miss sth or there was supposed to be 0 rather than -1 as the value of the part circled with red pen around 15:00?
Sorry for my unjustified cockyness :-/ Of course there’s supposed to be -1 there. Need to follow the thought process more carefully and don’t question the expert’s competence... Keep it up maestro :-D
Well you nailed it at the end, a lot of fun watching it.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : th-cam.com/video/3vAvpeqJkEs/w-d-xo.html
The limit of 1/x as x→∞ is Zero ... but 1/∞ is not zero, the limit of it is Zero. Meaning 1/∞ never gets to Zero, just arbitrarily close. 1/∞ is Asymptotically Equivalent to Zero. Asymptotic means approaching a value or curve arbitrarily closely (i.e., as some sort of limit is taken).
10:40 yeah because the other part was really easy to think of
OMG he wrote so many things with BLUE PEN.
could you expand the 2 square roots using the binomial theorem, get some cancellations when taking the difference, and somehow put a cap on the remainder?
Video length: twenty fuckin minutes
Me: isn't anything minus itself zero? And isn't zero to the power of anything, zero?
Thats me
My calc 2 teacher, I already graduated why am watching these (as in I have seen over a hundred) I guess masters here I come lol.
This is the one i thought Ramanujan and other math greats wouldn't have solved it
wouldnt the 2x just make the left a lot bigger than the right side? So the whole term goes to infinity?
Instead of using conjugated at 3:30 couldnt you just create an inequality to see which one grows faster?
the square root on the left grows faster because there's an extra +2x term.
But I'm not sure how you could conclude the entire thing is equal to one, just from that.
-lim = -1/2, so ln(L) is just 1/2 right?
An awesome limit I've ever seen. Love it.