Why $ \lim_{x\rightarrow\infty}\frac{\cos(x)-x}{x+2} $ doesn't work with L'Hopital?

Question

$$ \lim_{x\rightarrow\infty}\frac{\cos(x)-x}{x+2} $$

The limit is of the form $ \frac{-\infty}{\infty} $. Is this form not considered in the same way as $ \frac{\infty}{\infty} $?

If I differentiate numerator and denominator separately I get:

$$ \lim_{x\rightarrow\infty}\frac{-\sin(x)-1}{1} $$

However, this limit dances constantly between 0 and -2.

All L'Hôpital tells you is that, in the case of suitable indeterminates (as here), IF the limit of the derivatives exists then it equals the original limit. It is silent in cases (such as here) where the limit of derivatives fails to exist. — lulu, Aug 23 '21 at 16:18
Because this rule requires existence of $\lim \frac{f'}{g'}$. It states that if this limit exists (and some other technical conditions), then $\lim \frac{f}{g}$ exists and they are equal. — Yalikesifulei, Aug 23 '21 at 16:21
@DatBoi I wonder why this question was suddenly closed but the your previous one is still open. It seems there is not a uniform way to evaluate similar questions which deserve closure. — user, Aug 23 '21 at 16:50
@user it happens quite a lot, it depends on how the question was framed and the experience of the user answering it, etc, etc. How many times have you seen a question on an indefinite integral where OP thinks the answer is wrong just because it differs by a constant? :P(a meta post on this maybe?) — DatBoi, Aug 23 '21 at 17:09
See also https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-calculating-lim-x-to-infty-fracxx-s — Barry Cipra, Aug 23 '21 at 19:24

Oscar Lanzi · Answer 1 · 2021-08-23T20:05:38.357

2

L'Hopital's Rule says the initial and detivative-based limits are equal if the latter exists. Here the derivative-based limit does not exist, so we do not meet the conditions that enable L'Hopital's Rule.

There are many alternative methods. One is to squeeze the numerator between $1-x$ and $-(1+x)$, both of which numerators give derivative-based limits that exist and agree if we choose to apply L'Hopital's Rule. Therefore the limit is identified as this common value.

edited Aug 23 '21 at 20:05

answered Aug 23 '21 at 16:29

Oscar Lanzi

39,403

Actually, L'Hopital says the initial limit exists (and equals the derivative-based limit) provided the latter exists; if it required that you know both limits exist in order to say they're equal, it wouldn't be nearly as useful as it is. – Barry Cipra Aug 23 '21 at 19:28

score 1 · Answer 2 · answered Aug 23 '21 at 16:20

1

Not always l'Hospital works, indeed the theorem doesn't state that it works always, but of course we can proceed as follows

$$\frac{\cos(x)-x}{x+2}=\frac x x \frac{\frac{\cos(x)}x-1}{1+\frac 2x}$$

from which we can easily conclude.

answered Aug 23 '21 at 16:20

user

154,566

score 1 · Answer 3 · answered Aug 23 '21 at 16:26

As others have indicated, if the limit of $f/g$ is an indeterminate form and the limit of $f'/g'$ exists (or continue taking successive derivatives until you no longer reach an indeterminate form), then you can for sure conclude that the limit of $f/g$ exists and determine what its value is. It is not an if and only if. There are situations (like this one) where L'Hopital fails but the limit of the original function does exist. Here's a hint.

You can rewrite your function as:

$$ \frac{\cos(x)-x}{x+2} = \frac{\cos(x)+2-x-2}{x+2} = \frac{\cos(x)+2 - (x+2)}{x+2} = \frac{\cos(x)+2}{x+2} - 1.$$

score 1 · Answer 4 · answered Aug 23 '21 at 16:29

If a limit fulfils the condition to apply L Hopital (numerator and denominator are differentiable ) and after it's application the limit exists then only you can say that that's the answer to the limit. But, if the limit post application of L Hopital does not exist, you can say nothing...you need to find another way to compute the limit.

Why $ \lim_{x\rightarrow\infty}\frac{\cos(x)-x}{x+2} $ doesn't work with L'Hopital?

4 Answers4