TITLE: $\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$
QUESTION [12 upvotes]: Evaluate
$$S=\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$

Its quite simple to conclude that this limit does not exist.
Now here comes the interesting part. Consider the limit
$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{1-\frac{\sin x}x}{1+\frac{\cos x}x}=1$$
But this limit is of the form $\frac \infty\infty$. So from L'Hopital's
$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{(x-\sin x)'}{(x+\cos x)'}=S$$
$$\implies S=1$$

Is this reverse reasoning correct? Is there any faulty assumption that is made here?

REPLY [18 votes]: L'Hopital's rule says, in this case, that if the limit$$\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$exists, then it is equal to$$\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}.$$You tried to apply this implication in the wrong direction, which is not valid.