@@technicalengineer6300 technically you are correct but your exam probably won't like that because limit 1/x is "does not exist" x->0 in first place; so you cannot split lim
@@technicalengineer6300 mcq! why are you trying to solve it without cheating? i mean use a *calculator*, put the same formal but without lim and replace x with something very close to zero (for example 0.0000001 or -0.00000001) you will instantly get your results
@@Ninja20704it's an immediate consequence of the squeeze theorem, so to my knowledge it is "proper". if a function f(x) is bounded by two real numbers a and b then a
You should look up this function, it looks really funny, there is basically a bow tie around (0,0) because the line just goes up and down drawing 2 triangles
That's just 0 because the value of sin(1/X) will always be between -1 and 1 for all real values of X, so when you do the limit as X tends to 0 of X*sin(1/X) the same as saying 0*k where k is some vale between -1 and 1, therefore the limit is equal to 0
knowing myself, i just did something unholy.... i broke sine function into exponential complex form, then where we get e^(i/x) we basically get e^0 or 1, same goes for e^-(i/x), and we get (1-1)/2i uptop, and that's 0, and then x goes to zero, sine goes to zero so the whole thing goes to zero.... why did i do this
Could you leverage the fact that x and sin(1/x) are both odd, so the product is even? This would automatically make the left limit equal to the right limit since the left side is a reflection over the y-axis. I've never seen limit laws that involve odd or even functions though.
So... what "odd" means is that f(-x) = -f(x). And yeah, you could certainly observe that fact and use it to express the left-hand limit _in terms of_ the right-hand limit. Not sure that would have made this particular problem easier, though.
I did it a different way. I know that sin(1/x) is going to be rapidly oscillating between -1 and 1 and at 0 we have infinite oscillations between two FINITE values being multiplied by an infitesmially small value thus resulting in another infinitesimally small value. using the even-odd identity I know for a fact this proof works for both of the sides limits because it doesn’t matter if you go from 1 to -1 or -1 to 1 the oscillations still occur in that finite range and the x is still pushing the whole thing to 0. If x approached from the negative side rather than positive it’s the same as if it approached from the positive and the sine just had a negative 1 multiplied with it (even odd identity), in either case you get either an infinitely small value approaching 0 from either the negative or positive side. This means the limit is 0.
Sin(x) is always between 1 and -1 (given x is real), and all finite numbers will result in 0 when they are multiplied by 0. So when sin(x) of a large x is multiplied by 0, it will always equal 0. Hence, despite the fact sin(x) doesn't have a limit as x gets large, x*sin(1/x) will approach 0 as x approaches 0.
No. To play devil's advocate, we can consider the limit of x*tan(1/x) as x approaches zero. Tangent of anything is unbounded, and even approaches infinity near every odd multiple of pi/2. This means you get an indeterminate form if there is a possibility that tan(1/x) is near any odd multiple of pi/2. Because there are infinitely many of these forbidden values of the domain of tangent, you cannot conclusively say that tan(1/x) is finite as x approaches zero. You can only say that 0*L is 0, if you can conclusively lock down L to either be a finite number, or be bounded by finite numbers.
0•unbounded is indeterminate 0•bounded is 0 But what exactly does Ramanujan Principle Value Theorem state? Never heard of it and failed to look it up on the internet.
Hey! I've been trying to find an answer for a question we've come up with in high school, (but neither us nor our teachers could answer) and it is in connection with calculus. It may be trivial but her it is: Let's take a function f(x) (could be either real or complex valued we did not specify that), let's start taking the derivatives of f(x), like so f'(x), f''(x), f'''(x)...f'n(x) (where n goes to infinity). Is there an algorithm that can decide whether this will converge to a sort of loop (like sin(x)) or to a constant value like 0 or e^x or it will grow into a bigger and bigger expression (that is more complicated with each step), without actually calculating all the derivatives one by one and just looking at the starting function f(x)? So I'm basically asking whether this problem is decidable or undecidable (like Conway's game of life).
We just need an o(1) function (under the relevant topological base, such as x->0) multiplied by an O(1) function (under that same base). We do not need for the O(1) function to be bounded specifically by 1 and -1.
An alternative technique I thought of is by using the substitution u=1/x and rewriting the limit in terms of u: u=1/x => x=1/u, if the bound needs to be x -> 0 then 0 = 1/u and u must approach infinity. then the limit will be: limit{u -> infinity} (sin(u)/u) which is the popular cardinal sine function. and by Its graph one can see it goes to 0 when u -> infinity. thus the limit = 0
@@Playerofakind For every ball B(0, r) of positive real radius r, centered at 0, it is true that there is a punctured neighbourhood U(infinity)\{infinity}, such that for all u from that neighbourhood it is true that f(u) = sin(u)/u is in B(0, r), because for u > r we have |sin(u)/u| < |1/u| < 1/r.
* 한국어로 작성하여 죄송합니다 제 생각에는 sin x 의 함수의 평균값이 0 에 수렴하기 때문인것같습니다. sin 함수의 그래프는 원점에서 시작하고 또 1회의 주기에 대하여 그래프와 x축이 이루는 밑넓이의 값이 0이 됩니다. 1 /x 가 무한대로 가면서 그 함수값의 평균은 0에수렴합니다. 더불어. 무한소를 함수의 앞에 곱하므로 농도가 두배인 무한소가 만들어지며 우리는 이것을 x 에 비해 상대적 으로 0에근접하니 0 이라고 이해할수 있습니다. 제 해석이 고등학교 2학년수준에서 적합한 해석인지 평가해주셨으면 좋겠습니다.
@@Ninja20704Alternatively, we can consider lim x→0 |xsin(1/x)|. Note that 0≤|xsin(1/x)|≤|x|. Since lim x→0 |x|=0, by Squeeze Theorem we have lim x→0 |xsin(1/x)|=0. As the absolute limit tends to 0, then lim x→0 xsin(1/x)=0. We can use ε-δ language to prove that lim |f(x)|=0 implies lim f(x)=0
This is not about your video but please answer me... Hi, I'm from France and I'm super interested in mathematics! I have several questions, could you solve the following integrals: the integral of |((sin(x))/(x))| and the integral of 1/(1+x^(5)). I know these are big integrals but it would be nice to do them ! PS: can you make a complete guide for the 3rd year of calculus with your McDonald's employee outfit? I don't have more to say! Bye! :D
There is no elementary solution to the the general indefinite integral sin(x)/x dx. We have to define a special function called Si(x), which stands for sine integral, in order to do it. The definite integral from 0 to infinity of sin(x)/x is called the Dirichlet integral, and is equal to pi/2. One way to evaluate it, is to use the Laplace transform. The Laplace transform of sin(x)/x is arctan(1/s). Using the initial value theorem (limit as s goes to infinity), and final value theorem (s goes to zero), of arctan(1/s)/s, we can show that the integral evaluates to pi/2 across these limits. For 1/(x^5 + 1), there is an elementary solution. You can factor the denominator as (x^4 - x^3 + x^2 - x + 1)*(x + 1), to get one of the solutions. Then you can use the quartic formula to factor the quartic, which has all complex solutions. From that point on, it's just integration by partial fractions. I believe BPRP has a video on this one.
Still using the squeeze theorem, can we avoid having to split into a left half and a right half by just noticing that xsin(1/x) is always between -|x| and |x|
I found an easier way than the squeeze theorem. make a u-sub: let u = 1/x so the limit becomes: Iim u --> infinity (1/u * sin u) but since sin u is always between -1 and 1 and also lim u --> infinity (1/u) = 0 we conclude that: lim x-->0 (x*sin(1/x)) = 0.
What's the point of this substitution? After the substitution you did everything the same way BPRP did, and you still use the squeeze theorem implicitly when you conclude that the relevant limit is 0 after noting that -1
Surely you can just say that its equal to 0 by looking at it? In any case, plugging in 0 that function will give you zero since what ever sin(1/x) is, it exist 0 times. Additionally, is sin(∞) not just oscillating between -1 and 1, so that you wont have an infinite value, in that case you wont have 0 times ∞, am i missing something here?
'plugging in 0 that function will give you zero' It won't, because 1/x is not defined for x=0. Furthermore, this would only be helpful if the function was continuous at x=0, and you would need to prove that first. 'Additionally, is sin(∞) not just oscillating between -1 and 1' It's obviously not defined on that point, and if it was defined as a function with the codomain [-1, 1], its value at that point would also be a single point from the line segment [-1, 1], and it wouldn't be 'oscillating'. 'am i missing something here?' You are defining a bunch of things your own way and forget to prove additional stuff, like whether or not x*sin(1/x) is continuous given your definitions.
please correct me if i'm wrong here, lim x->0 (sinx)/x is equal to 1. So if we re write the ques in video from xsin(1/x) to lim x->0 (sin(1/x))/1/x, shouldnt that be equal to 1 as well?
I substituted x for 1/x which makes the term sin(x)/ x . But the limit is now x→∞. Thus the denominator gets infinitely large but the numerator is bounded between 1 and -1. Hence the value tends to zero
@@wearron But sine(1/x) is from the original question. I intended to use a limit of sin(x)/x as x approaches infinity. And since in our case x is approaching 0, then 1/x approaches infinity, so we can sub t = 1/x, and t approaches inf. Where is the issue in my reasoning?
@@wearronsin(1/x) being indeterminate does not matter much, it is more important to note that sin(1/x) is bounded. Intuitively we have bounded/infinity=0.
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0:0 ("undefined" vs "indeterminate") calculus basics
th-cam.com/video/IohnkapKvvQ/w-d-xo.html
I think it is easy just take (x) in denometer and it become (1/x) so it is now lim…sin(1/x)/(1/x)
@@technicalengineer6300
technically you are correct but your exam probably won't like that
because limit 1/x is "does not exist"
x->0 in first place; so you cannot split lim
@@theplant4046 my exam pattern is mcq
Please, I want you to create a formula to solve the equation
ax^3+bx^2+cx+b=0
And this one too
ax^3+bx=n
@@technicalengineer6300
mcq! why are you trying to solve it without cheating?
i mean use a *calculator*, put the same formal but without lim and replace x with something very close to zero (for example 0.0000001 or -0.00000001)
you will instantly get your results
My reasoning was that regardless of what x is, sin(anything) is between -1 and 1, so you have 0*[-1..1] which will be 0.
Sure that’s the easy way but the proper way to do it is the squeeze theorem like he showed.
Use The Fundamental Theorem of Engineering Part 4. We get sinx = x
0 * x is 0 so the answer is 0.
@@_elusivex_ well that would unsolvable even at the engineering world, 0.infinity dne
@@Ninja20704it's an immediate consequence of the squeeze theorem, so to my knowledge it is "proper". if a function f(x) is bounded by two real numbers a and b then a
@@_elusivex_that doesn't work bc it's sin(1/x)
The sinx=x thing only works when x is very small, but 1/x becomes huge as x approaches 0
Hello I am the person who asked this question! Thank u for this video! Definitely really helpful
Glad to help and wishing you all the best!
I like how you have taken the time to answer questions for people and you don't talk down to them.
You should look up this function, it looks really funny, there is basically a bow tie around (0,0) because the line just goes up and down drawing 2 triangles
That's just 0 because the value of sin(1/X) will always be between -1 and 1 for all real values of X, so when you do the limit as X tends to 0 of X*sin(1/X) the same as saying 0*k where k is some vale between -1 and 1, therefore the limit is equal to 0
very nice video - makes me remember when I could barely solve such a question. Good times.
Love these videos where he answers random questions from reddit. Brings me back to my Calc 1 & 2 days.
The graph is pretty nice
0*bounded is always 0 by pinching theorem
Pinching Theorem is also called the Squeeze Theorem, which is used in the video.
I Used this function as an examples of path which is not rectifiable. Yesterday in my Advanced Complex Analysis class.
knowing myself, i just did something unholy.... i broke sine function into exponential complex form, then where we get e^(i/x) we basically get e^0 or 1, same goes for e^-(i/x), and we get (1-1)/2i uptop, and that's 0, and then x goes to zero, sine goes to zero so the whole thing goes to zero.... why did i do this
Because Steve taught you so well, that you are thinking in complex numbers.
@@carultch lmao
Could you leverage the fact that x and sin(1/x) are both odd, so the product is even? This would automatically make the left limit equal to the right limit since the left side is a reflection over the y-axis. I've never seen limit laws that involve odd or even functions though.
So... what "odd" means is that f(-x) = -f(x). And yeah, you could certainly observe that fact and use it to express the left-hand limit _in terms of_ the right-hand limit. Not sure that would have made this particular problem easier, though.
There isn't really a need, because we can squeeze this function just fine
@@thetaomegathetait's not about need, it's just exploration of another approach
@@chrisjfox8715
That approach seems to be the same thing, just with extra steps.
Yea, it doesn't make the solution "better". I just mention it because it's a nice intuition ✔️.
You can take 1/x as t. Then t approaches + or - infinity, make cases and solve
Amazing😍😍. We actually say this theorem as Sandwich theorem, as it is in between f(x) and g(x)
Squeeze theorem and sandwich theorem are the same
Weirdly, going straight to the definition of a limit, delta=epsilon just works. 0
The Squeeze theorem using bounded sides
Same as: lim u -> inf sin(u)/u = 0 quite obviously
0:11 looks like squeeze theorem
I did it a different way. I know that sin(1/x) is going to be rapidly oscillating between -1 and 1 and at 0 we have infinite oscillations between two FINITE values being multiplied by an infitesmially small value thus resulting in another infinitesimally small value.
using the even-odd identity I know for a fact this proof works for both of the sides limits because it doesn’t matter if you go from 1 to -1 or -1 to 1 the oscillations still occur in that finite range and the x is still pushing the whole thing to 0.
If x approached from the negative side rather than positive it’s the same as if it approached from the positive and the sine just had a negative 1 multiplied with it (even odd identity), in either case you get either an infinitely small value approaching 0 from either the negative or positive side.
This means the limit is 0.
Which app and which device you use to make such videos?
Sin(x) is always between 1 and -1 (given x is real), and all finite numbers will result in 0 when they are multiplied by 0. So when sin(x) of a large x is multiplied by 0, it will always equal 0.
Hence, despite the fact sin(x) doesn't have a limit as x gets large, x*sin(1/x) will approach 0 as x approaches 0.
yeah that's what he did in the video. It's called the squeeze theorem
the answer is easy but solving it is hard
Isn't 0•DNE=0 always, by Ramanujan principle value theorem?
No. To play devil's advocate, we can consider the limit of x*tan(1/x) as x approaches zero.
Tangent of anything is unbounded, and even approaches infinity near every odd multiple of pi/2. This means you get an indeterminate form if there is a possibility that tan(1/x) is near any odd multiple of pi/2. Because there are infinitely many of these forbidden values of the domain of tangent, you cannot conclusively say that tan(1/x) is finite as x approaches zero.
You can only say that 0*L is 0, if you can conclusively lock down L to either be a finite number, or be bounded by finite numbers.
0•unbounded is indeterminate
0•bounded is 0
But what exactly does Ramanujan Principle Value Theorem state? Never heard of it and failed to look it up on the internet.
Hey! I've been trying to find an answer for a question we've come up with in high school, (but neither us nor our teachers could answer) and it is in connection with calculus. It may be trivial but her it is: Let's take a function f(x) (could be either real or complex valued we did not specify that), let's start taking the derivatives of f(x), like so f'(x), f''(x), f'''(x)...f'n(x) (where n goes to infinity). Is there an algorithm that can decide whether this will converge to a sort of loop (like sin(x)) or to a constant value like 0 or e^x or it will grow into a bigger and bigger expression (that is more complicated with each step), without actually calculating all the derivatives one by one and just looking at the starting function f(x)? So I'm basically asking whether this problem is decidable or undecidable (like Conway's game of life).
So it seems that the same method can be used for x, x², x³ etc multiplied by and trig function that varies between 1 and -1?
We just need an o(1) function (under the relevant topological base, such as x->0) multiplied by an O(1) function (under that same base). We do not need for the O(1) function to be bounded specifically by 1 and -1.
An alternative technique I thought of is by using the substitution u=1/x and rewriting the limit in terms of u:
u=1/x => x=1/u, if the bound needs to be x -> 0 then 0 = 1/u and u must approach infinity.
then the limit will be: limit{u -> infinity} (sin(u)/u) which is the popular cardinal sine function. and by Its graph one can see it goes to 0 when u -> infinity.
thus the limit = 0
Doesn't the limit of sin(u)/u not exist since it's always oscillating as you tend to infinity? Maybe I misread
@@Playerofakindit exists because the function slowly approaches the x axis, and this can be proven with the squeeze theorem too.
@@Playerofakind
For every ball B(0, r) of positive real radius r, centered at 0, it is true that there is a punctured neighbourhood U(infinity)\{infinity}, such that for all u from that neighbourhood it is true that f(u) = sin(u)/u is in B(0, r), because for u > r we have |sin(u)/u| < |1/u| < 1/r.
very nice
* 한국어로 작성하여 죄송합니다
제 생각에는 sin x 의 함수의 평균값이 0 에 수렴하기 때문인것같습니다. sin 함수의 그래프는 원점에서 시작하고 또 1회의 주기에 대하여 그래프와 x축이 이루는 밑넓이의 값이 0이 됩니다. 1 /x 가 무한대로 가면서 그 함수값의 평균은 0에수렴합니다. 더불어. 무한소를 함수의 앞에 곱하므로 농도가 두배인 무한소가 만들어지며 우리는 이것을 x 에 비해 상대적 으로 0에근접하니 0 이라고 이해할수 있습니다.
제 해석이 고등학교 2학년수준에서 적합한 해석인지 평가해주셨으면 좋겠습니다.
I wonder how would you tackle this DE.
y' = (x (2-3x)) / (3y^2 - 1)
can we write it as sin(1/X)/1/X...then lim =1 right?
I thought the same thing
This is equal to 1 only if 1/X tends to 0. Many students just remember the form and the result, but don’t remember the requirements.
@@pneujai tnx
I feel like you could take a bidirectional limit and not need to repeat the same step right?
No you can’t. Because we cannot multiply the inequality by x unless we break down into two cases x>0 and x
@@Ninja20704Alternatively, we can consider lim x→0 |xsin(1/x)|.
Note that 0≤|xsin(1/x)|≤|x|.
Since lim x→0 |x|=0, by Squeeze Theorem we have lim x→0 |xsin(1/x)|=0.
As the absolute limit tends to 0, then lim x→0 xsin(1/x)=0.
We can use ε-δ language to prove that lim |f(x)|=0 implies lim f(x)=0
I would use epsilon delta to prove it's 0
This is not about your video but please answer me...
Hi, I'm from France and I'm super interested in mathematics! I have several questions, could you solve the following integrals: the integral of |((sin(x))/(x))| and the integral of 1/(1+x^(5)). I know these are big integrals but it would be nice to do them ! PS: can you make a complete guide for the 3rd year of calculus with your McDonald's employee outfit? I don't have more to say! Bye! :D
There is no elementary solution to the the general indefinite integral sin(x)/x dx. We have to define a special function called Si(x), which stands for sine integral, in order to do it.
The definite integral from 0 to infinity of sin(x)/x is called the Dirichlet integral, and is equal to pi/2. One way to evaluate it, is to use the Laplace transform. The Laplace transform of sin(x)/x is arctan(1/s). Using the initial value theorem (limit as s goes to infinity), and final value theorem (s goes to zero), of arctan(1/s)/s, we can show that the integral evaluates to pi/2 across these limits.
For 1/(x^5 + 1), there is an elementary solution. You can factor the denominator as (x^4 - x^3 + x^2 - x + 1)*(x + 1), to get one of the solutions. Then you can use the quartic formula to factor the quartic, which has all complex solutions. From that point on, it's just integration by partial fractions. I believe BPRP has a video on this one.
Still using the squeeze theorem, can we avoid having to split into a left half and a right half by just noticing that xsin(1/x) is always between -|x| and |x|
Sure it's also correct.
I found an easier way than the squeeze theorem.
make a u-sub:
let u = 1/x
so the limit becomes: Iim u --> infinity (1/u * sin u)
but since sin u is always between -1 and 1 and also lim u --> infinity (1/u) = 0
we conclude that:
lim x-->0 (x*sin(1/x)) = 0.
What's the point of this substitution? After the substitution you did everything the same way BPRP did, and you still use the squeeze theorem implicitly when you conclude that the relevant limit is 0 after noting that -1
you still need to use the squeeze theorem after the substitution, this substitution is pointless
Sir please solve my problem integration of x^n tanx ??🙏
Do it 'by parts'
@@raja2850 sir nahi hota please 🥺 nikal do na
@@raja2850 sir plese solve
@@arunmalha7510 n ka value hai ya nahi?
Sir n ka value nahi hai
Tempted by the truth of the limit.
Tempted but the truth is discover, now that zero's done
There's no answer...
Surely you can just say that its equal to 0 by looking at it? In any case, plugging in 0 that function will give you zero since what ever sin(1/x) is, it exist 0 times. Additionally, is sin(∞) not just oscillating between -1 and 1, so that you wont have an infinite value, in that case you wont have 0 times ∞, am i missing something here?
'plugging in 0 that function will give you zero'
It won't, because 1/x is not defined for x=0.
Furthermore, this would only be helpful if the function was continuous at x=0, and you would need to prove that first.
'Additionally, is sin(∞) not just oscillating between -1 and 1'
It's obviously not defined on that point, and if it was defined as a function with the codomain [-1, 1], its value at that point would also be a single point from the line segment [-1, 1], and it wouldn't be 'oscillating'.
'am i missing something here?'
You are defining a bunch of things your own way and forget to prove additional stuff, like whether or not x*sin(1/x) is continuous given your definitions.
@@thetaomegatheta Thanks for the response:) I do see my mistake.
Sandwich theorem
why to bother about sides of zero, when sine is odd function??
Wheres the pokeball
A bprp vid WITHOUT THE WHITEBOARD
Doesnt feel right
I use the sandwich rule for these
Bounded times zero is trivially zero. Easy to show by εδ
Can I, instead, just say that "since sine is always between -1 and 1, and zero times anything between -1 and 1 is zero, then the result must be zero"?
Yes. More precisely, there is a theorem
If lim f = 0 and g is bounded, then lim fg = 0.
please correct me if i'm wrong here, lim x->0 (sinx)/x is equal to 1. So if we re write the ques in video from xsin(1/x) to lim x->0 (sin(1/x))/1/x, shouldnt that be equal to 1 as well?
I substituted x for 1/x which makes the term sin(x)/ x . But the limit is now x→∞. Thus the denominator gets infinitely large but the numerator is bounded between 1 and -1. Hence the value tends to zero
i got 1 somehow
This problem is when x is approaching 0. I think you did when x approaches inf.
@@bprpcalculusbasics ty that clarified my mistake
white board? 😔
Oscillating, not classic DNE 👀
Well, this limit very much does exist in R, let alone in the two-point compactification of R.
Squeeze theorem? I like to call it the "sandwich theorem"
I solved the zero minus case with symmetry because x sin(1/x) is an even function
Can we multiply it by (1/x)/(1/x)? Then we get sin(1/x)/(1/x) and since x approaches 0, 1/x approaches infinity so we get 0?
no, since sine(1/x) would be "undefined" or does not exist but in this limit is is zero, and also, . that's like saying 0/0 = 0
@@wearron But sine(1/x) is from the original question. I intended to use a limit of sin(x)/x as x approaches infinity. And since in our case x is approaching 0, then 1/x approaches infinity, so we can sub t = 1/x, and t approaches inf. Where is the issue in my reasoning?
@@adikac7081 There's no issue, it works fine as well :)
Yes you can, but to be formal, you need to use the Squeeze Theorem like in the video too. So actually it makes the whole thing redundant.
@@wearronsin(1/x) being indeterminate does not matter much, it is more important to note that sin(1/x) is bounded. Intuitively we have bounded/infinity=0.
Or you can just ignore the value of x at all cuz:
X × sin × (1/x) → sin (x/x) → sin (1) → 0
x*sin(1/x) is not equal to sin(x/x).