$\lim_{x\rightarrow \infty}\frac{x+\cos x}{x+\sin x}$ Show that L'Hôpital's rule gives no answer. Compute $$\lim_{x\rightarrow \infty}\frac{x+\cos x}{x+\sin x}$$ by common sense. Show that L'Hôpital's rule gives no answer.
I saw this question required the use of common sense. L'Hôpital's rule does not help but how is that supposed to be shown/proved?
Do you have some hints?
 A: If you calculate the derivative of the numerator and the denumerator, then you have $$\lim_{x \to \infty} \dfrac{1-\sin x}{1+\cos x},$$ and this limit does not exist, since $\lim_{x \to \infty} \sin x$ and $\lim_{x \to \infty} \cos x$ do not exist.
However, $\lim_{x \to \infty} \dfrac{x+\cos x}{x+\sin x} = \lim_{x \to \infty} \dfrac{1+\frac{\cos x}{x}}{1+ \frac{\sin x}{x}} = 1$
A: To show that L' Hopital's Rule doesn't work, just apply it. You get $\lim_{x  \to \infty} \frac{1 - \sin x}{1 + \cos x}$. Can you work that limit out? It may help to use a tangent half-angle substitution (Weierstrass substitution) to simplify that to a simple algebraic expression in terms of $\tan \frac x2$. Then it's even easier to see.
Remember L'Hopital's Rule depends on certain conditions. One of the conditions is the "reduced" limit actually exists. In this case, it doesn't.
To actually, work out the limit, the comments have already given you some great hints. Basically a simple bounding argument would work on top and bottom.
