L'Hopital's Rule seemingly not giving graphical result

Question

I'm trying to evaluate the following limit... $$\lim\limits_{x \to \infty} \frac{x-3}{\cos x+x}$$

My first thought was to use L'Hôpital's rule. The numerator obviously tends towards infinity, and so, I believe, does the denominator, because of the additive $x$ term. Applying the rule, then, we arrive at the following result: $$\lim\limits_{x \to \infty} \frac{1}{-\sin x+1}$$ It would therefore seem that the limit does not exist, since the denominator is an alternating trig function with periodic asymptotes. However, if we graph our original function, it's quite clear that the limit actually tends to 1. This can be shown be squeezing the function between $\frac{x-3}{-1+x}$ and $\frac{x-3}{1+x}$ respectively.

Why, then, does L'Hôpital's rule not seem to apply to this function? Is there some nuance or precondition for the rule that I'm missing?

Answer 1

L'Hopital's Rule assumes that $\lim \frac {f'(x)} {g'(x)}$ exists and then proves that $\lim \frac {f(x)} {g(x)}$ exist and the two limit are equal. The rule is not applicable here.

Answer 2

L'Hôpital's rule states that, for limits of the form $0/0$ and $\infty/\infty$, we have: $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)},$$ if the latter limit exists. In this case it does not, and so L'Hôpital's rule does not help us to reach any conclusions.