When to apply l'Hôpital rule?

Question

I'm confused about when to apply l'Hôpital Rule, here is an example:
\begin{align} L=\lim_{x \to\infty} \frac{1}{x^2}\int_0^x \frac{t}{2+\sin(t)}dt\\ \end{align}
Approach 1:
Let $F(x)$ is antiderivative of $\frac{x}{2+\sin(x)}$, then $\int_ 0^x \frac{t}{2+\sin(t)}dt= F(x)-F(0)= F(x)+C$. So we got our limit:
\begin{align} L=\lim_{x\to\infty} \frac{F(x)+C}{x^2}\\ \end{align}
It's easy to see that this is $\frac{\infty}{\infty}$ case so we apply l'Hôpital Rule:
\begin{align} L=\lim_{x\to\infty} \frac{\frac{d}{dx}(F(x)+C)}{\frac{d}{dx}(x^2)}=\lim_{x\to\infty} \frac{\frac{x}{2+\sin(x)}}{2x}= \lim_{x\to\infty} \frac{1}{4+2\sin(x)}\\ \end{align}
However, the limit results in nothing.
Approach 2:
Use $\lim _{x \to\infty} \frac{1}{x^2}\int_0^x tf(\sin(t))dt=\frac{1}{4\pi}\int_{-\pi}^{\pi} f(\sin(t))dt $. We have:
\begin{align} L= \frac{1}{4\pi}\int_{-\pi}^{\pi} \frac{1}{2+\sin(t)}dt=\frac{1}{2\sqrt{3}}\\ \end{align}
Can you help me point out any mistakes in approach 1 and why the l'Hôpital Rule doesn't work?

Answer 1

In order to apply the L Hospital's rule to find $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ when $\lim_{x\to\infty} g(x)=+\infty$ and $f,g$ are differentiable on $(0,\infty)$, it is necessary that $g'(x)\neq 0$ for all $x>a$ for some $a>0$ and, the limit $$\lim_{x\to\infty}\frac{f'(x)}{g'(x)}\text{ exists}$$ But here the limit $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$ does not exist and hence we can't use the L Hospital's rule.