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

Question

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

$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{(x-\sin x)'}{(x+\cos x)'}=S$$

$$\implies S=1$$

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

Answer 1

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.