TITLE: Why doesn't L'hopitals Rule work for $\lim\limits_{x \to \infty} \frac{x+ \sin x}{x+ 2 \sin x}$?
QUESTION [9 upvotes]: This is how I would evaluate $\lim\limits_{x \to \infty} \dfrac{x+ \sin x}{x+ 2 \sin x}$
$=\lim\limits_{x \to \infty} \dfrac{x \left( 1+ \frac{\sin x}{x} \right)}{x \left(1+ 2 \cdot \frac{ \sin x}{x} \right)}$
$= \dfrac{1+0}{1+2 \cdot 0} = 1$
But now applying L'hopitals Rule, I get
$\lim\limits_{x \to \infty} \dfrac{1+ \cos x}{1+ 2 \cos x}$ 
Since $\cos x $ just oscillates between $[-1,1]$ I think we can  conclude the limit doesn't exist.
What is going on here?

REPLY [2 votes]: Because your function doesn't satisfy the hypothesis. If you are studing the limit $x\to c$, in order to apply the theorem the function $g=x+2\sin x$ must be differentiable and $g'(x)\ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+\infty) where $g' \ne 0$. But $g'(x)=0$ $\forall x=-\frac{\pi}{4}+2k\pi$, so it doesn't exists a set like that.