College math during lectures: **Taught us simple derivative and l'hopital's rule problems** Me: This is fairly easy and fun! College math during exams: **Gave us every tricky l'hopital's rule problem that existed out there** Me: *FML*
This is great L'Hospital's rule practice, but if you're going for speed, the best way to approach the problem is to assume sin(x)=x, which is true for very small values of x like we have here. Then you can set the limit expression (now x^(1/ln(x))) equal to y, take the natural log of both sides, then use some log rule manipulation to get ln(y)=ln(x)/ln(x), which solves to y=e. Great L'Hospital's video regardless.
˜You can avoid some head scratching at the end by first raising ln(x) = ln(sinx)/ln(x) to the e power. You cancel log_e with e and you get x = exp(ln(sinx)/ln(x)) You can evaluate the exp from there
What's the problem with using the small angle approximation for sinx (i.e. sin(x)=x for small x in radians)? That way when you take ln of both sides, you get ln(y)=ln(x)/ln(x) = 1 so y approaches e. I guess since you go to the trouble of doing it this way, that you can't just use the small angle approximation, but why not?
8:53 Why we can't apply rule hear if by putting 0, we get 0/0? (x*cosx => 0*cos0 = 0*1 = 0 & sinx => sin0 = 0, so 0/0). By using L'Hopital's rule I get y=1
You can apply L'Hopital's rule as many times as you want so long as you are applying it to something that meets the requirements (limits that approach 0/0 or infinity/infinity). As he shows in this video, you often need to manipulate the limit to get it into one of those forms so you can use the rule, and you often need to use it more than once in the same problem.
No u cannot apply L'Hopital's rule as many times u want.because u reach a certain point where after applyinng derivative to the denominator u get zero.It means tthat we have to stop applying derivatives and then calculate limit of the final function
Praneeth Lanka That's true for polynomials, but for something like sin(x) or cos(x), no matter how many derivatives I take I never get 0. On the other hand, yes, if I started with two polynomials and took enough derivatives, I'd get 0.
Thanks for this. Really helpful. I wish you could get the colors of your pens right, though and be well prepared for every video before you shoot, cause it's very distracting and quite annoying when you keep writing and erasing stuff every second
Surely a sign of this generation. You have a MIT graduate, world-class tutor teaching you, for free, and you can do it on demand, watch during the morning, or at night or whenever you feel like it, pause it, re-wind, whatever. You are the first generation of humans to have this luxury. And you are complaining - about the ink colors the tutor is using because he’s not switching them quickly enough for you. Mind you, he could do it all in white like a chalk board, but he’s going the extra step of putting items in different colors to make it even clearer for you. And, despite this luxury that not even a prince or princess could get, you are complaining.
Wouldn't it be much easier to solve it like this: lim(x->0) (sinx)^1/lnx can be rewrited as lim(x->0) e^ln((sinx)^1/lnx) (because e^x and ln(x) are inverse functions to each other). Now, beacuse we know e^f(x) is always increasing, we can put the limit right before the exponent, therefore: lim(x->0) (sinx)^1/lnx = e^lim(x->0) ln((sinx)^1/lnx). For now we can forgote the e and calculate only lim(x->0) ln((sinx)^1/lnx). We can put the exponent in front of the ln => lim(x->0) 1/lnx * ln(sinx) => lim(x->0) (ln(sinx))/(lnx) and we are when you were. Now we can use l'hopital rule and simply get, that lim(x->0) ln((sinx)^1/lnx) = 1. Now we substitute 1 back to the exponent of the e and we get our result as e^1 = e. I think this is much easier and shorter solution not for only this limit, but for limits in the form something^something.
The use of L'Hopital's Rule to find the limit of x/sinx is can't be justified since in order to find the derivative of sinx you must evaluate the limit of [sinx]/x which is just the reciprocal of x/sinx.
is it because one uses the fundamental limit idea to prove the derivative of sin(x) itself? So meaning that using derivative of sin(x) to find lim (sinx/x) would be incorrect?
There are some nice geometrical ways to prove the derivative of sin(x) without that limit, but I agree that he should've referenced his old geometric proof of the limit instead. The logic isn't technically circular, but with the information that he's taught it sure looks circular, so it's not the best way to present the solution.
As the limit approaches 0, Sinx/x=1..... Which is the same thing as sinx/x=1/1....so x/sinx (as limit approaches 0) is 1..... No need to to use L'hopital's rule at this stage.
instead of applying one last time the rule, could i just inverse x/sinx to (sinx/x)^-1 and then that would mean that with lim x -> 0 ((sinx/x)^-1) = (1)^-1 = 1 ?
Ivan Žagar I'm not so sure it would be the same, since the lim as x -> 0 in this case is 0 from the positive direction. Anyhow, if the limit is indeed x -> 0 and you wish to make use of (sinx)/x as x -> 0, then there's no need to inverse (inverse of 1 is 1).
Yo hold on. If you didn't separate the limit into two parts, you could still use L'Hôspital and find the answer on (x*cos x)/sin x as 1. Why is this possible? Please answer me before my head explodes.
WAIT of zero times,infinity equals zero,andnthe limit of cos x/sin x equals infity when you plug in zero..couldn't you say the limit is one times zero and therefore zero? Where's the flaw?
I know I’m a bit late, but the issue is that these are all indeterminate forms. 0/0 or inf/inf cases can evaluate to essentially anything. The whole idea of calc is essentially to get things out of indeterminate forms.
No, 0^0 is undefined. In very simple terms you can think of it as possibly being anything because if you take another limit than the one provided in this example(for example the limit as x approaches 0 of sin(x) ^ x starts out as the same 0 ^ 0 undefined case but the value of the limit is 1), the value of that limit is not necessarily e even though at first you had the same 0 ^ 0 undefined case.
Makes it kinda magical to me. You really shouldn't be able to just make it come out however you want. If these kind of functions appear in physics I would be alarmed. ;)
wresing But, you're not really making it come out how you want. In this case, the 0^0 expression HAS to come out e. In another case, it may HAVE to come out to something else. But if the limit does exist, then that's the way it has to be. I'm sure there are countless cases in physics where L'Hopital's rule is used and concrete results are extracted from indeterminate forms.
wresing You can't make it come out whatever you want for a fixed function but you could construct a function such that 0 ^ 0 ends up being whatever value you wish(say you want ( + / - ) a, a > 0, then a possible function to give you this value if you take the limit of it as x approaches 0 is (+/-) * [ sin(x) ^ ( ln(a) / ln(x) ) ] .
Try to evaluate the following limits as they go to 0: {2^(-1/x)}^x {10^(-1/x)}^x {e^(-1/x)}^x {ㅠ^(-1/x)}^x {[2^(1/2)]^(-1/x)}^x You'll very quickly see why mathmaticians get angry when you say 0^0 equal to something.
took 5 mins tops, Which terrifies me because you say it's tricky but when compared to my teacher's exercises It's way easier. I'm so screwed o_o edit: after reading some coments, it seems tricky is a slight overstatement or missunderstood by me as difficult.
College math during lectures: **Taught us simple derivative and l'hopital's rule problems**
Me: This is fairly easy and fun!
College math during exams: **Gave us every tricky l'hopital's rule problem that existed out there**
Me: *FML*
THIS
What is going on.
Lmao
😂
This is great L'Hospital's rule practice, but if you're going for speed, the best way to approach the problem is to assume sin(x)=x, which is true for very small values of x like we have here. Then you can set the limit expression (now x^(1/ln(x))) equal to y, take the natural log of both sides, then use some log rule manipulation to get ln(y)=ln(x)/ln(x), which solves to y=e. Great L'Hospital's video regardless.
L'Hopital (no S) is a French mathematician.
@@siphilipe yea they are refering to a video about l'hopital's rule not about him clearly.
As my wise friend says, "Fck around and you'll find an answer in Math". Awesome
Hats off! U can make a complicated problem seem so easy and creative! Thanx!
I was really young when this video was made and today it proved to be helpful to me
Thanks alot sir
You could have just raised the whole function e^ln since f(x)=e^lnf(x) and solved from there. I feel like it takes less time and is easier.
+Ian Mayle True, but this was a more interesting way to solve it.
actually he tries to demonstrate a fact that implies on other functions in general
So much fun!!!! (sarcasm). Will be happy when I graduate college and never see this again. Thanks for the video though.
Higher maths and the proofs are way more fun 🙂
Ka Gg : "Yeah say that to the other masochist"
it is kinda fun
It's actually fun( no sarcasm)😁
˜You can avoid some head scratching at the end by first raising ln(x) = ln(sinx)/ln(x) to the e power. You cancel log_e with e and you get x = exp(ln(sinx)/ln(x)) You can evaluate the exp from there
Dear god I suck at this. Thank you for this vid
Nice work, great explanation. Keep up the good work (Y)
What's the problem with using the small angle approximation for sinx (i.e. sin(x)=x for small x in radians)? That way when you take ln of both sides, you get ln(y)=ln(x)/ln(x) = 1 so y approaches e.
I guess since you go to the trouble of doing it this way, that you can't just use the small angle approximation, but why not?
Solved in 1 minute...you need to redefine tricky but thanks for the challenge
4:33 is it because the number e appears in nature? Also the term "NATURAL log." Green is a color we associate with nature.
this is awesome
The example: lim x->0 sinx/x
The test:
but the thing is, this problem isn’t too hard, unlike, say the extreme algebra question on bprp’s video, which is a far better example of this.
1 lol
Who ever got this at the first get go, you're a genius in my world. I am really bad at math.
Nice solving! That's great.
It is very simple, I have done it in high school and university
Never thought maths was actually fun
Wow this is fun
Can you tell me the tool that you use to write these examples? Because I want to apply it to enhance my learning
No offence but i think that its not a tricky problem rather it is a basic one. Great explaination by the way. Thumbs up from me. 👍👍☺️☺️
8:53 Why we can't apply rule hear if by putting 0, we get 0/0? (x*cosx => 0*cos0 = 0*1 = 0 & sinx => sin0 = 0, so 0/0). By using L'Hopital's rule I get y=1
Yeah I got 1 as well
You can apply it but you would have to use product rule, so to make it less complicated he factored out cos(x)
nice video even though it's kinda confusing
Interesting. really.
thats so cool
This was so entertaining! It made me laugh when the answer was e at the end. What an interesting problem the think about.
more problems please
Can you apply L'Hopital's rule as many times as you want until you get and appropriate limit?
No, check the wikipedia page(for example) to see when you can apply L'Hopital's rule.
You can apply L'Hopital's rule as many times as you want so long as you are applying it to something that meets the requirements (limits that approach 0/0 or infinity/infinity). As he shows in this video, you often need to manipulate the limit to get it into one of those forms so you can use the rule, and you often need to use it more than once in the same problem.
No u cannot apply L'Hopital's rule as many times u want.because u reach a certain point where after applyinng derivative to the denominator u get zero.It means tthat we have to stop applying derivatives and then calculate limit of the final function
Praneeth Lanka That's true for polynomials, but for something like sin(x) or cos(x), no matter how many derivatives I take I never get 0. On the other hand, yes, if I started with two polynomials and took enough derivatives, I'd get 0.
As long as the form is indeterminate you can apply it
Whew I got it right! 😅
Thanks for this. Really helpful. I wish you could get the colors of your pens right, though and be well prepared for every video before you shoot, cause it's very distracting and quite annoying when you keep writing and erasing stuff every second
Surely a sign of this generation. You have a MIT graduate, world-class tutor teaching you, for free, and you can do it on demand, watch during the morning, or at night or whenever you feel like it, pause it, re-wind, whatever. You are the first generation of humans to have this luxury.
And you are complaining - about the ink colors the tutor is using because he’s not switching them quickly enough for you. Mind you, he could do it all in white like a chalk board, but he’s going the extra step of putting items in different colors to make it even clearer for you. And, despite this luxury that not even a prince or princess could get, you are complaining.
Wouldn't it be much easier to solve it like this: lim(x->0) (sinx)^1/lnx can be rewrited as lim(x->0) e^ln((sinx)^1/lnx) (because e^x and ln(x) are inverse functions to each other). Now, beacuse we know e^f(x) is always increasing, we can put the limit right before the exponent, therefore: lim(x->0) (sinx)^1/lnx = e^lim(x->0) ln((sinx)^1/lnx). For now we can forgote the e and calculate only lim(x->0) ln((sinx)^1/lnx). We can put the exponent in front of the ln => lim(x->0) 1/lnx * ln(sinx) => lim(x->0) (ln(sinx))/(lnx) and we are when you were. Now we can use l'hopital rule and simply get, that lim(x->0) ln((sinx)^1/lnx) = 1. Now we substitute 1 back to the exponent of the e and we get our result as e^1 = e. I think this is much easier and shorter solution not for only this limit, but for limits in the form something^something.
The use of L'Hopital's Rule to find the limit of x/sinx is can't be justified since in order to find the derivative of sinx you must evaluate the limit of [sinx]/x which is just the reciprocal of x/sinx.
+colton ellis hmmm why so?
To do so is a use of circular logic.
is it because one uses the fundamental limit idea to prove the derivative of sin(x) itself?
So meaning that using derivative of sin(x) to find lim (sinx/x) would be incorrect?
There are some nice geometrical ways to prove the derivative of sin(x) without that limit, but I agree that he should've referenced his old geometric proof of the limit instead. The logic isn't technically circular, but with the information that he's taught it sure looks circular, so it's not the best way to present the solution.
As the limit approaches 0, Sinx/x=1..... Which is the same thing as sinx/x=1/1....so x/sinx (as limit approaches 0) is 1..... No need to to use L'hopital's rule at this stage.
At 0 sin = 0, I don;t know what you're getting at
So, the graph of this function does not look like e is the limit. Should it?
Pls Short trick video for integral solution to prepare Indian navy SSR
i beg to differ, differentiating ln(sin x)/ ln(x) you should have used quotient rule but you have differentiated directly.
It was a easy question but u made it feel difficult. Especially for us (IITIAN'S)
try to bring a hard question next time.
wth is the difference between log and ln
This should be pretty easy to solve if you ever wondered and solved for yourself what the derivative of x ^ x is.
MrIStillDontCare yup ;)
Hospitals' rule sounds easier to do.
instead of applying one last time the rule, could i just inverse x/sinx to (sinx/x)^-1 and then that would mean that with lim x -> 0 ((sinx/x)^-1) = (1)^-1 = 1 ?
Ivan Žagar I'm not so sure it would be the same, since the lim as x -> 0 in this case is 0 from the positive direction. Anyhow, if the limit is indeed x -> 0 and you wish to make use of (sinx)/x as x -> 0, then there's no need to inverse (inverse of 1 is 1).
Yeah, it’s valid but an unnecessary step.
I solved this easily... and I am the very ordinary kid in class.. is that allowed..?
No, it’s not allowed. Please turn over your mathematics license.
r u an angel
What are the demerits or precautions to use LH rule
Don't use it
The question was damn easy!!!!
anyone know what kind of tablet he is using ?
or what pen to computer interface he is using?
I Want to use it to do my homework in one note.
Can anyone tell me what is that software he is using in the video
instead of x/sinx can u use x * cot x
1:26 Why does the natural log of values as they get closer to zero approach neative infinity? I thought it doesn't exist becasuse ln 0 doesn't exist.
Properties of natural log. X^-infinity = x/ infinity. Which is equal to zero. This means that Ln 0 = - infinity
Thanks for the video but the problem was not that tricky if we use only standard limits and solve and do not use L'Hopitals Rule
Change the white board colour we can not see or understand on black colour
or just change sin(x) to x at 5:49 hehe
Hi. So that last step; you just took e of both sides right? (e of lny and e of 1)
There’s no such thing as “taking e of both sides.”
that is the most beautiful mouse-writing words I ever seen before XD.
he is using a wacom tablet sorry to break the truth to u though
Using l'hôpital to calculate lim_(x→0)(x/sin(x)) is kinda heretic
hey is the derivative taken in a wrong manner. i mean diff of a/b is not diff of a/ diff of b.
He is not taking the derivative of the whole thing.
He is applying the L'Hopital rule
The answer is e
Can anyone solve this problem?
Limit (1+b/x)^x/a
X approches infinity
Per the binomial theorem, it’s e^(b/a)
Yo hold on.
If you didn't separate the limit into two parts, you could still use L'Hôspital and find the answer on (x*cos x)/sin x as 1. Why is this possible?
Please answer me before my head explodes.
xcosx/sinx is as x-->0 is the indeterminant form 0/0 so you could use L'Hopitals rule again.
7:58 why the quotient rule was not applied?
Because you’re not taking the derivative of the whole thing. You’re differentiating the top and the bottom.
I thought it was pronounce as L hospital
WAIT of zero times,infinity equals zero,andnthe limit of cos x/sin x equals infity when you plug in zero..couldn't you say the limit is one times zero and therefore zero? Where's the flaw?
I know I’m a bit late, but the issue is that these are all indeterminate forms. 0/0 or inf/inf cases can evaluate to essentially anything. The whole idea of calc is essentially to get things out of indeterminate forms.
Does this mean 0^0=e?
No, 0^0 is undefined.
In very simple terms you can think of it as possibly being anything because if you take another limit than the one provided in this example(for example the limit as x approaches 0 of sin(x) ^ x starts out as the same 0 ^ 0 undefined case but the value of the limit is 1), the value of that limit is not necessarily e even though at first you had the same 0 ^ 0 undefined case.
Makes it kinda magical to me. You really shouldn't be able to just make it come out however you want. If these kind of functions appear in physics I would be alarmed. ;)
wresing But, you're not really making it come out how you want. In this case, the 0^0 expression HAS to come out e. In another case, it may HAVE to come out to something else. But if the limit does exist, then that's the way it has to be. I'm sure there are countless cases in physics where L'Hopital's rule is used and concrete results are extracted from indeterminate forms.
wresing
You can't make it come out whatever you want for a fixed function but you could construct a function such that 0 ^ 0 ends up being whatever value you wish(say you want ( + / - ) a, a > 0, then a possible function to give you this value if you take the limit of it as x approaches 0 is (+/-) * [ sin(x) ^ ( ln(a) / ln(x) ) ] .
Try to evaluate the following limits as they go to 0:
{2^(-1/x)}^x
{10^(-1/x)}^x
{e^(-1/x)}^x
{ㅠ^(-1/x)}^x
{[2^(1/2)]^(-1/x)}^x
You'll very quickly see why mathmaticians get angry when you say 0^0 equal to something.
Hahaha let me write it in “language” 😂🍤
Sir it con be translated in hindi
❤
Lol I am just 16 why am I learning this now
is this tricky..??
oh, change colour, change colour, change colour... unnecessary...
Min height in CDS exam
This guy definitely have OCD
took 5 mins tops, Which terrifies me because you say it's tricky but when compared to my teacher's exercises It's way easier. I'm so screwed o_o edit: after reading some coments, it seems tricky is a slight overstatement or missunderstood by me as difficult.
Dunnit
Uh, what
Ohh
not even tricky
This was really juicy 😋 😩 👌
Θα πεσει πανελληνιες
по русски говори