2
$\begingroup$

I'm having a hard time evaluating the following limit

$$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}$$ I'm not even really sure how to approach it since I'm not used to seeing another function in the integral.

My initial thoughts were to use l'hopitals rule since $\lim_{x \rightarrow0} 2\sin x=0$, so the numerator is equal to 0, but I wouldnt be sure how to take the derivative of the numerator.

Any pointers would be appreciated

$\endgroup$
  • $\begingroup$ Try using L'Hospital rule followed by the Leibniz rule $\endgroup$ – Rohan Shinde Aug 16 '18 at 2:01
  • $\begingroup$ Thanks for the tip. Do you think there is any other way to approach besides liebniz rule? I dont think we've learned liebniz rule in my calc 1 class, but this is from an old problem set, so the course may have changed. $\endgroup$ – Butts Carlton Aug 16 '18 at 2:06
0
$\begingroup$

We will use that $$f(x)=\int_0^{a(x)}g(t)dt\implies f'(x)=g(a(x))a'(x).$$ In our case:

$$\frac{d}{dx} \int_{0}^{2 \sin x} \cos(t^2) dt=\cos((2\sin x)^2)\cdot \dfrac{d(2\sin x)}{dx}.$$

So, we have

$$\frac{d}{dx} \int_{0}^{2 \sin x} \cos(t^2) dt=2\cos (x)\cos(4\sin^2 x).$$

Thus

$$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0}\dfrac{2\cos (x)\cos(4\sin^2 x)}{2} =1.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do you mind elaborating on your first step? I'm able follow after that but $$\frac{d}{dx} \int_{0}^{2 sinx} cos(t^2) dt=\cos((2\sin x)^2)\cdot \dfrac{d(2\sin x)}{dx}.$$ is unclear to me. Thank you $\endgroup$ – Butts Carlton Aug 16 '18 at 2:12
  • $\begingroup$ I have editted the answer to clarify it. Is it clear now? $\endgroup$ – mfl Aug 16 '18 at 2:13
  • $\begingroup$ Yes, thank you. I wasn't thinking about the integral as a composition. $\endgroup$ – Butts Carlton Aug 16 '18 at 2:20
2
$\begingroup$

Without Liebnitz rule, you can use first mean value theorem for integrals, then there exist $c$ such that $$\lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0} \frac{2 \sin x \cos(c^2) }{2x}=\lim_{x \rightarrow 0} \frac{2 \sin x }{2x}\lim_{x \rightarrow 0} \cos(c^2)=\lim_{x \rightarrow 0} \cos(c^2)=1$$ because with sandwich $0\leq c\leq2\sin x$ so $c\to0$ as $x\to0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The denominator $2x$ can be replaced by $2\sin x$ via the limit $\lim_{x\to 0}(\sin x) /x=1$ and then putting $u=2\sin x$ the limit is easily seen to be $$\lim_{u\to 0} \frac{1}{u}\int_{0}^{u}\cos(t^2)\,dt=\cos(0^2)=1$$ via Fundamental Theorem of Calculus.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Alternatively, use Taylor expansion: $$\cos(t^2)=1-\frac{t^4}{2}+\frac{t^8}{24}+O(t^{9});\\ \int_{0}^{2\sin x} \cos(t^2)dt=\left(t-\frac{t^5}{10}+O(t^6)\right)\bigg{|}_0^{2\sin x}=2\sin x-\frac{(2\sin x)^5}{10}+O((2\sin x)^6);\\ \lim_{x \rightarrow 0} \frac{\int_{0}^{2 \sin x} \cos(t^2) dt}{2x}=\lim_{x \rightarrow 0}\frac{2\sin x+O(\sin^5x)}{2x}=\lim_{x \rightarrow 0}\frac{2\sin x}{2x}\cdot (1+O(\sin^4x))=1.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.