Why $ \lim_{x\rightarrow\infty}\frac{\cos(x)-x}{x+2} $ doesn't work with L'Hopital? $$ \lim_{x\rightarrow\infty}\frac{\cos(x)-x}{x+2} $$
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.
 A: 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.
A: 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.
A: 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:
$$ \frac{\cos(x)-x}{x+2} = \frac{\cos(x)+2-x-2}{x+2} = \frac{\cos(x)+2 - (x+2)}{x+2} = \frac{\cos(x)+2}{x+2} - 1.$$
A: 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.
