# Why doesn't L'hopitals Rule work for $\lim\limits_{x \to \infty} \frac{x+ \sin x}{x+ 2 \sin x}$? [duplicate]

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?

## marked as duplicate by Zacky, Namaste, Pierre-Guy Plamondon, Eric Wofsey, Lord Shark the UnknownDec 28 '18 at 2:47

• See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently. – Zacky Dec 27 '18 at 13:50
• This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + \cos x)/x$. – Ben Dec 27 '18 at 13:51
• – Barry Cipra Dec 27 '18 at 14:06
L'Hospital's rule contains an assumption that $$\lim_{x \to a} f'(x)/g'(x)$$ exists, which is not true in this case.
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.
• The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $\;3x+2\sin x\;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ... – DonAntonio Dec 27 '18 at 14:06