$\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$

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?

• "this limit does not exist since the limits $\cos(x)$ and $\sin(x)$ do not exist". The limit indeed does not exists, but this reasoning is wrong. Consider, for instance $\lim_{x \to \infty} \sin(x)/\sin(x)$. Mar 4 at 6:54
• @Magdiragdag I understand. I just meant it in an informal way Mar 4 at 6:56

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.