limit $\lim_{x\rightarrow 0^{+}}\frac{1}{x^2}\cdot \int^{x}_{\sin(x)}\frac{1}{\sqrt{1+\sin(u)}}du$

Question

Evaluation of $$\lim_{x\rightarrow 0^{+}}\frac{1}{x^2}\cdot \int^{x}_{\sin(x)}\frac{1}{\sqrt{1+\sin(u)}}du$$

What I try

Applying D L'Hospital's Rule,

$$\lim_{x\rightarrow 0^+}\frac{1}{x}\bigg(\frac{1}{\sqrt{1+\sin x}}-\frac{\cos x}{\sqrt{1+\sin(\sin x)}}\bigg)$$

Again Limit is in $\displaystyle \frac{0}{0}$ form

So $$\lim_{x\rightarrow 0+}\frac{1}{4}\bigg[-(1+\sin x)^{-\frac{3}{2}}\cdot \cos (x)+\cos (x)\cdot (1+\sin(\sin x))^{-\frac{3}{2}}\cos(\sin x)\cdot \cos(x)-2\sin x(1+\sin(\sin x))^{-\frac{1}{2}}=0$$

Is my answer is Right.

Or any other short way by which we can solve it, please explain me

Answer 1

If you know Taylor's formulas / $O$ notation, this is as usual faster and less painful than L'Hospital. Informally, since you are looking at $x$ small, the integrand is essentially 1, so the limit is the same as $$ \frac{x - \sin x}{x^2}, $$ and as $x - \sin x = O(x^3)$, then this tends to $0$. To be slightly more precise, for $u \in (0,\pi/2)$, $\sin u \geq 0$, so the integrand is bounded by 1, and therefore, for $x \in (0,\pi/2)$, $$ 0 \leq \frac{1}{x^2} \int_{\sin x}^{x} \frac{1}{\sqrt{1+\sin u}} \: \mathrm{d}u \leq \frac{x - \sin x}{x^2}, $$ and conclude as before.