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?

  • 3
    $\begingroup$ "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, 2021 at 6:54
  • $\begingroup$ @Magdiragdag I understand. I just meant it in an informal way $\endgroup$
    – DatBoi
    Mar 4, 2021 at 6:56

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.