# How to find limit of sin^2(x)/x^2 as x--> infinity

Hello Everyone,
I'm doing some practice problems and I am stuck on the following limits.
1) Limit as x approaches infinity of (sin^2(x))/x^2
Now for this one I thought that the answer would just be 1 seeing as the limit of sinx/x is 1, however the answer is apparently 0 and I can't see the reasoning for this.
2) Limit as x approaches 0 of (cosx-1)/xsinx
Now for this one I first tried getting an the identity for sin^2x out of it by squaring it but that didn't work. Then I just tried breaking it up, but I just keep ending up getting 0 but apparently the answer is 1/2.
3) Limit as x approaches pi/3 of (cosx-1/2)/(x-pi/3)
Again I am totally lost. I keep wanting to try and get an identity out of it but I just cant get anything useful.
4) Now this question I know should be easy. I am just having trouble understanding where to start. If F(x) = the integral of f(t)dt (the values on the integral are b=x and a=1) and f(t)= integral (squareroot(1+u^2))/u *du . Find F''(3).
Apparently the answer to this is (2squareroot82)/3
If this isn't clear I can always draw it out and post it, I'm sure that what I tryed to type isn't overly clear.
Sorry about these questions guys, I know I haven't exactly shown a lot of work, but these are the ones that I have no idea of how to do, I've been attempting them for 2 days now and just can't seem to get anywhere

Dale
Mentor
2020 Award
scorpa said:
1) Limit as x approaches infinity of (sin^2(x))/x^2
Now for this one I thought that the answer would just be 1 seeing as the limit of sinx/x is 1, however the answer is apparently 0 and I can't see the reasoning for this.
The limit of $$\frac{\sin (x)}{\Mfunction{x}}$$ is only 1 as x approaches 0, not as x approaches infinity. This problem is unrelated to that one.

The key to this one is to realise that $$\frac{{\sin (x)}^2}{x^2}$$is always between $$\frac{1}{x^2}$$ and 0.

I think you can use L'hopital's (sp?) rule for 2 and 3.

-Dale

Last edited:
Hmmm ok, the way I interpreted that problem was in the form of (sinx)(sinx)/x^2 is this wrong? I always thought that sin^2x =(sinx)(sinx) and sinx^2/x^2 =$$\frac{{\sin (x)}^2}{x^2}$$

Dale
Mentor
2020 Award
$${\sin (x)}^2 = \sin (x)\,\sin (x) = {\sin }^2 (x)$$
Mathematica just uses the notation that I used while most textbooks favor your notation.

Either way, you are looking at the wrong part, in one case (sinx/x) you have the limit as x approaches 0, in the other (this problem) you have the limit as x approaches infinity. The squares have nothing to do with it in this case.

-Dale

Last edited:
Oh ok I see, my mistake I see what you are getting at now. I am still having trouble with the other questions though, I just don't see where to start :(

Dale
Mentor
2020 Award
scorpa said:
I am still having trouble with the other questions though, I just don't see where to start :(
2 and 3 are just standard L'hopital's rule problems. Just apply his rule until you don't get 0/0.

I didn't really follow 4. I might be able to help if you could TEX it, but I might still be lost.

-Dale

I will try and write it out and paste a link to number 4. As for 2 and 3 I dont actually know that rule so I will look it up on the net and see what I can come up with from that. Thanks

Ok so with the L'Hopitals rule, all you have to do is take the derivative of the numerator and denominator and keep doing it until you don't get 0/0 anymore>???? Does this always work like this? Can I use this rule all the time in a situation like this? Based on reading about the rule this is how I solved the questions and I ended up with the right answer:

2) cox-1/xsinx ---> -sinx/(sinx-xcosx)---> -cosx/cosx+cosx-xsinx which when you substitute in x=0 gives you an answer of -1/2!!!!

3) (cosx-1/2)/(x-pi/3) -----> -sinx--->-sin(pi/3)---> -sqrt3/2!!!!

If I can use that method all the time that is awesome! I never knew about it before, in class we were never shown that!!!

Thanks a lot!!!!

HallsofIvy
Homework Helper
scorpa said:
If $$F(x)= \int_1^x f(t)dt$$ and $$f(t)= \int_1^{t^2} \frac{1+ u^2}{u}du[/itex], find F'(3)$$

Do you know the "Fundamental Theorem of Calculus": If $$F(x)= \int_1^x f(t)dt$$ the F'(x)=f(x). In particular, $$F'(3)=f(3)= \int_1^{9} \frac{1+ u^2}{u}du[/itex]. To do that integral, multiply both numerator and denominator by u to get [tex]F'(3)= f(3)= \int_1^{9} \frac{u(1+ u^2)}{u^2}du[/itex], and let v= 1+ u^2. JasonRox Homework Helper Gold Member scorpa said: Ok so with the L'Hopitals rule, all you have to do is take the derivative of the numerator and denominator and keep doing it until you don't get 0/0 anymore>???? Does this always work like this? Can I use this rule all the time in a situation like this? Based on reading about the rule this is how I solved the questions and I ended up with the right answer: 2) cox-1/xsinx ---> -sinx/(sinx-xcosx)---> -cosx/cosx+cosx-xsinx which when you substitute in x=0 gives you an answer of -1/2!!!! 3) (cosx-1/2)/(x-pi/3) -----> -sinx--->-sin(pi/3)---> -sqrt3/2!!!! If I can use that method all the time that is awesome! I never knew about it before, in class we were never shown that!!! Thanks a lot!!!! See you were never shown that in class, I would recommend to only use it as a test. Your teacher probably expects you to find it through another route. HallsofIvy said: Do you know the "Fundamental Theorem of Calculus": If [tex]F(x)= \int_1^x f(t)dt$$ the F'(x)=f(x). In particular, $$F'(3)=f(3)= \int_1^{9} \frac{1+ u^2}{u}du[/itex]. To do that integral, multiply both numerator and denominator by u to get [tex]F'(3)= f(3)= \int_1^{9} \frac{u(1+ u^2)}{u^2}du[/itex], and let v= 1+ u^2. Where does the square root go in that integral? And how do you get the second derivative from that? Stupid questions I know, but I'm just now seeing this one . And how would you go about solving the limit problems without L'hopitals rule? Now that I learned it I rather like it....it's very handy! Dale Mentor 2020 Award scorpa said: Ok so with the L'Hopitals rule, all you have to do is take the derivative of the numerator and denominator and keep doing it until you don't get 0/0 anymore>???? Does this always work like this? Can I use this rule all the time in a situation like this? There are (of course) some limitations. If you are trying to take the limit of some f(x)/g(x) then the most important limitation is that the limit of f'(x)/g'(x) must exist. For example, let [tex]f(x) = \sin (x)$$ and $$g(x) = x$$ then $$\Mfunction{Limit}(\frac{f(x)}{g(x)}, {x\rightarrow {\infty}}) = 0$$. However, $$f'(x) = \cos (x)$$ and $$g'(x) = 1$$ so $$\Mfunction{Limit}(\frac{f'(x)}{g'(x)}, {x\rightarrow {\infty}})$$ does not exist. So for this one L'Hopital's rule doesn't work. Instead you have to use the same method as in problem 1 to get the right answer.

-Dale

Last edited:
VietDao29
Homework Helper
DaleSpam said:
There are (of course) some limitations. If you are trying to take the limit of some f(x)/g(x) then the most important limitation is that the limit of f'(x)/g'(x) must exist.
For example, let $$f(x) = \sin (x)$$ and $$g(x) = x$$ then $$\Mfunction{Limit}(\frac{f(x)}{g(x)}, {x\rightarrow {\infty}}) = 0$$. However, $$f'(x) = \cos (x)$$ and $$g'(x) = 1$$ so $$\Mfunction{Limit}(\frac{f'(x)}{g'(x)}, {x\rightarrow {\infty}})$$ does not exist. So for this one L'Hopital's rule doesn't work. Instead you have to use the same method as in problem 1 to get the right answer.
-Dale
???
What is this???
What do you mean???
---------------------
L'Hopital rule is used when both numerator and denominator either tends to 0 or infinity, i.e:$$\frac{0}{0} \quad \mbox{or} \quad \frac{\infty}{\infty}$$. It states that, if:
$$\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}$$ exists, and either:
$$\lim_{x \rightarrow \alpha} f(x) = 0 \quad \mbox{and} \quad \lim_{x \rightarrow \alpha} g(x) = 0$$ or:
$$\lim_{x \rightarrow \alpha} f(x) = \infty \quad \mbox{and} \quad \lim_{x \rightarrow \alpha} g(x) = \infty$$, then:
$$\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}$$.
-------------------
#1, you cannot use L'Hopital rule since you don't know what sin2(x) tends to when x tends to infinity. It can be done using Squeeze theorem, by noticing the fact that:
$$0 \leq \frac{\sin ^ 2 x}{x ^ 2} \leq \frac{1}{x ^ 2}$$
-------------------
#2, you have two different ways to do it without using L'Hopital rule:
The first way is to use power-reduction formulas:
$$\sin ^ 2 x = \frac{1 - \cos(2x)}{2}$$.
So here you have:
$$\lim_{x \rightarrow 0} \frac{\cos x - 1}{x \sin x} = \lim_{x \rightarrow 0} - \frac{2 \sin ^ 2 \left( \frac{x}{2} \right)}{x \sin x} = \lim_{x \rightarrow 0} - \frac{2 \sin ^ 2 \left( \frac{x}{2} \right)}{4 \frac{x}{2} \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}$$.
Can you go from here?
The second way is to multiply both numerator and denominator by (1 + cos x), so that you'll have sin2x in the numerator.
It goes like this:
$$\lim_{x \rightarrow 0} \frac{\cos x - 1}{x \sin x} = \lim_{x \rightarrow 0} \frac{(\cos x - 1)(1 + \cos x)}{x \sin x (1 + \cos x)} = \lim_{x \rightarrow 0} \frac{- \sin ^ 2 x}{x \sin x (1 + \cos x)}$$.
Can you go from here?
-------------------
#3, you can change 1 / 2 into: $$\frac{1}{2} = \cos \left( \frac{\pi}{3} \right)$$, then use the identity:
$$\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)$$. Can you go from here?
L'Hopital rule is overkill... and as a matter of fact, you haven't learnt it, so you shouldn't use it.
--------------------
If $$F(x) := \int_{\alpha} ^ {x} f(t) dt$$ then $$F'(x) = \frac{d \left( \int_{\alpha} ^ {x} f(t) dt \right)}{dx} = f(x)$$. There should be a proof in your book, or you can view it here (see part 3).
So $$F''(x) = f'(x) = \frac{d \left( \int_{1} ^ {x ^ 2} f(u) du \right)}{dx} \ = \ ?$$, where $$f(u) := \frac{\sqrt{1 + u ^ 2}}{u}$$. Note that the upper limit of the integral is x2, not x.
Can you go from here?

Last edited:
OK, thanks so much! I think I understand those limits now even without that handy little rule. And I can now get the answer for that integral! Thanks very much guys!