# evaluating $\lim_{x \rightarrow 0} \frac{\int_{0}^{2\sin x} \cos(t^2) dt}{2x}$

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

• Try using L'Hospital rule followed by the Leibniz rule – Rohan Shinde Aug 16 '18 at 2:01
• 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. – Butts Carlton Aug 16 '18 at 2:06

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

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

• 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 – Butts Carlton Aug 16 '18 at 2:12
• I have editted the answer to clarify it. Is it clear now? – mfl Aug 16 '18 at 2:13
• Yes, thank you. I wasn't thinking about the integral as a composition. – Butts Carlton Aug 16 '18 at 2:20

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 $x\to0$.

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.

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