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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?

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    $\begingroup$ "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 $\lim_{x \to \infty} \sin(x)/\sin(x)$. $\endgroup$ Mar 4 at 6:54
  • $\begingroup$ @Magdiragdag I understand. I just meant it in an informal way $\endgroup$
    – DatBoi
    Mar 4 at 6:56
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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.

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