Find $$\lim_{x \to \infty} \frac{x-\sin(x)}{x - \cos(x)}$$

It can be easily solved by dividing both numerator and denominator by $x$ leading to answer as $1$.

Now, my doubt is, since it is of the form $\frac{\infty}{\infty}$ , I can use L'Hopital's Rule here after which, it leads to

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

Now, according to me, this limit does not exist because $\sin(x)$ and $\cos(x)$ can be anything in $[-1,1]$ at $\infty$.

Where am I going wrong?

  • $\begingroup$ As you say the limit is easily solved, so why do you even think about using the Hospital? $\endgroup$ – Angina Seng Nov 8 '17 at 19:01
  • $\begingroup$ You can't use $$\lim_{x\to\infty}\frac{1-\cos(x)}{1+\sin(x)}=\frac{\lim_{x\to\infty} 1-\cos(x)}{\lim_{x\to\infty} 1+\sin(x)}$$ $\endgroup$ – Dave Nov 8 '17 at 19:01
  • 3
    $\begingroup$ Hospital says that if the limit of the quotient of the derivatives exists, then the limit of the functions do and they are equal. Emphases on the "if" there. $\endgroup$ – David Mitra Nov 8 '17 at 19:02
  • $\begingroup$ The question has an answer here math.stackexchange.com/questions/2500171/… $\endgroup$ – Rene Schipperus Nov 8 '17 at 19:13

Very simple: L'Hospital's rule can be applied under two conditions, which are often forgotten:

  • $f'(x)/g'(x)$ must have a limit,
  • $g'(x)$ must not vanish in some neighbourhood of the point at which the limit is to be calculated (except at the point itself).

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