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Consider the Limit $\frac{x+\sin x}{x-\sin x}$ as $x \rightarrow \infty$. Clearly this limit exists and is equal to $1$. Also notice that the function is of the form $\frac{\infty}{\infty}$ and thus, L'Hopitals rule can be applied here. But on applying L'Hopitals rule, I am getting Limit $\frac{1+\cos x}{1-\cos x}$ as $x \rightarrow \infty$, which is obviously non-converging. Where did I go wrong?

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    $\begingroup$ L'Hospital's Rule doea not require the numerator limit to be infinity when the denominator limit is infinity. But that is not the issue here. The limit of the quotient of derivatives needs to exist. It does not and LHR is inapplicable. $\endgroup$
    – Mark Viola
    May 28 at 16:32
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You cannot apply L'Hospital's rule because $\dfrac{1+\cos(x)}{1-\cos(x)}$ does not have a limit when $x \rightarrow \infty$. The existence of this limit is an essential hypothesis to apply L'Hospital's rule.


Instead you can factorize by $x$,

$$\dfrac{x+\sin(x)}{x-\sin(x)} =\dfrac{x(1+\sin(x)/x)}{x(1-\sin(x)/x)} = \dfrac{1+\sin(x)/x}{1-\sin(x)/x}$$ and the limit is $1$ as you pointed out.

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  • $\begingroup$ Is there some kind of deeper meaning behind this situation where we get a non-existent limit on applying L'Hopitals rule? $\endgroup$ May 28 at 17:05
  • $\begingroup$ @AarthBhardwaj I never use L'Hospital's rule, I think we can always find other ways to calculate limits. Anyway for me the "deeper meaning" is that the limit's existence of the ratio $f'/g'$ is fundamental to prove this rule. This is not perfectly rigorous, but to give you the gist, the proof uses squeeze theorem by giving an upper and a lower bound of the terms of $f/g$ with the terms of $f'/g'$ and if this ratio does not have a limit, for example if it oscillates like in your example, then you cannot "squeeze" $f/g$ using $f'/g'$. $\endgroup$
    – Axel
    May 28 at 17:33
  • $\begingroup$ @AarthBhardwaj For more informations, you can check: en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$
    – Axel
    May 28 at 17:34

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