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I know how to solve this using the squeeze theorem, but I am supposed to solve only using L'Hôpital's rule

$$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi}$$

I tried: $$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi} = \lim\limits_{x \to \infty} \frac{d/dx[\sin(x)]}{d/dx[x-\pi]} = \lim\limits_{x \to \infty} \frac{\cos(x)}{1}$$

From here I am stuck because the rule no longer applies and using $\infty$ for $x$ doesn't not help to simplify.

Logically the limit is $0$ because $\sin(x)$ can only be $-1$ to $1$, but this is using squeeze theorem.

Is still there any way to solve this without using the squeeze theorem?

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    $\begingroup$ L'Hopital's rule cannot be used, as $\frac{\sin x}{x-\pi}$ is not an indeterminate form as $x\to\infty$. $\endgroup$ – carmichael561 Jan 4 '17 at 22:10
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    $\begingroup$ Are you sure that $x \to \infty$? My bet is that $x \to \pi$, and that $\infty$ in your question is a typo. $\endgroup$ – Alex M. Jan 4 '17 at 22:15
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    $\begingroup$ I feel like this question doesn't deserve a down vote it's clear that OP took time to write this question and put his own thoughts.It's not his fault that there is probably a typo in his book. $\endgroup$ – kingW3 Jan 4 '17 at 22:26
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    $\begingroup$ I second @kingW3, and I also don't understand the close vote. The question is reasonable, and the OP's first on Math.SE, so let's not be that harsh. $\endgroup$ – Alex M. Jan 4 '17 at 22:28
  • $\begingroup$ Despite what everyone's saying here, Hospital can be used when the denominator has infinite limit (it doesn't matter how the numerator behaves). It fails here, because the limit of the quotient of the derivatives doesn't exist. $\endgroup$ – David Mitra Jan 4 '17 at 22:40
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Unfortunately L'Hopital would not help since this does not satisfy the assumptions.

Note that $$ \left|\frac{\sin x}{x-\pi}\right|\leq\frac{1}{|x-\pi|}\to 0 $$ as $x\to\infty$.

What is the point not to use the squeeze theorem?

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    $\begingroup$ But this is the squeeze theorem... since one needs to know that $|\text{anything}|\ge0$. $\endgroup$ – Simply Beautiful Art Jan 4 '17 at 22:11
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I bet that your problem is, in fact, to compute

$$\lim _{x \to \color{red} \pi} \frac {\sin x} {x - \pi} ,$$

so either you have mistyped $\infty$ instead of $\pi$, or there is a typo in the text where you took this from.

In this case, L'Hospital's theorem could be used, but it's not necessary, because

$$\lim _{x \to \pi} \frac {\sin x} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - 0} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - \sin \pi} {x - \pi} = (\sin ') (\pi) = \cos \pi = -1 .$$

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  • $\begingroup$ I double checked, and the text I took it from still had it as approaching infinity. This was probably a book typo. $\endgroup$ – C.A Jan 4 '17 at 22:44
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    $\begingroup$ @A.C: If the book explicitly requires the use of l'Hospital's theorem then it is surely a typo. If the problem was just included among others solvable with this theorem, or if it followed the chapter discussing l'Hospital's theorem, then there is a slight possibility that it was intended to check whether students pay attention to the hypotheses of a problem, or just apply a result blindly. In other words, there is a slight possibility that l'Hospital's theorem was a "red herring" in this problem. We'll probably never know. $\endgroup$ – Alex M. Jan 4 '17 at 22:53
  • $\begingroup$ (+1) unlikely that $x\to \infty$ $\endgroup$ – Mark Viola Jan 5 '17 at 3:43
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L'Hopital's Rule only works when the limit is $0/0$ or $\infty/\infty$. In this case, the limit is undefined over $\infty$, so you cannot use L'Hopital's Rule on this problem (at least the way it's written.)

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  • $\begingroup$ Your statement is false. See the note at the end of this section. If the other conditions of LHR are satisfied and the denominator approaches $\infty$, then LHR is applicable - even if the limit of the numerator fails to exist. Here, LHR is inapplicable since the limit of the quotient of derivatives fails to exist. $\endgroup$ – Mark Viola Jan 5 '17 at 3:45
  • $\begingroup$ Interesting. I didn't know that. I don't remember that being stated in any of the classes I've taken. $\endgroup$ – Nathan H. Jan 5 '17 at 18:43
  • $\begingroup$ It's often not discussed since if $g\to \infty$ and $f\to L$, then LHR really isn't necessary to find the limit of $f/g$. It becomes interesting only when either $f\to \infty$, which is the case you referenced, or when $\lim f$ fails to exist. $\endgroup$ – Mark Viola Jan 5 '17 at 18:55
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You can't. To apply L'Hospital's rule, it must be indeterminate form i.e. $\frac\infty\infty$ or $\frac00$.

As it is neither, you should simply use squeeze theorem.

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    $\begingroup$ Why would anyone downvote this? I won't upvote to compensate for the downvote, I dislike doing this, but I really can't understand what's wrong with this answer. $\endgroup$ – Alex M. Jan 4 '17 at 22:17
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    $\begingroup$ Actually, l’Hôpital can also be used in the case “whatever$/\infty$”; the problem here is that the limit after differentiating does not exist. You can see a note at en.wikipedia.org/wiki/L%27Hôpital%27s_rule#General_proof $\endgroup$ – egreg Jan 4 '17 at 22:24
  • $\begingroup$ @egreg Haha, I suppose. That is somewhat funny to think about :D $\endgroup$ – Simply Beautiful Art Jan 4 '17 at 22:25
  • $\begingroup$ @egreg: ...Which means that you can't use it (to compute limits, that is). $\endgroup$ – Alex M. Jan 4 '17 at 22:26
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    $\begingroup$ @AlexM. The curious thing is that the same answer (with a difference in the timestamp of about 30 seconds) got an upvote. The theorem can be used, but not in this case; the reason is different from what Simple Art states. $\endgroup$ – egreg Jan 4 '17 at 22:30

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