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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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L'Hospital's rule contains an assumption that $\lim_{x \to a} f'(x)/g'(x)$ exists, which is not true in this case.

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    $\begingroup$ Unfortunately, students often forget the hypotheses for a result when they try to apply it. $\endgroup$ – GEdgar Dec 27 '18 at 13:51
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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.

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  • $\begingroup$ 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 ... $\endgroup$ – DonAntonio Dec 27 '18 at 14:06
  • $\begingroup$ Thank you for the explanation! $\endgroup$ – ecrin Dec 27 '18 at 14:09

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