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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 $x\to\infty$. $\endgroup$ Jan 4, 2017 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, 2017 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, 2017 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, 2017 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. $\endgroup$ Jan 4, 2017 at 22:40

4 Answers 4

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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, 2017 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, 2017 at 22:53
  • $\begingroup$ (+1) unlikely that $x\to \infty$ $\endgroup$
    – Mark Viola
    Jan 5, 2017 at 3:43
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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 $|\text{anything}|\ge0$. $\endgroup$ Jan 4, 2017 at 22:11
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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, 2017 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, 2017 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, 2017 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, 2017 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, 2017 at 22:24
  • $\begingroup$ @egreg Haha, I suppose. That is somewhat funny to think about :D $\endgroup$ Jan 4, 2017 at 22:25
  • $\begingroup$ @egreg: ...Which means that you can't use it (to compute limits, that is). $\endgroup$
    – Alex M.
    Jan 4, 2017 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, 2017 at 22:30

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