# Why cannot I use L'Hospital Rule here?

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?

• As you say the limit is easily solved, so why do you even think about using the Hospital? – Angina Seng Nov 8 '17 at 19:01
• You can't use $$\lim_{x\to\infty}\frac{1-\cos(x)}{1+\sin(x)}=\frac{\lim_{x\to\infty} 1-\cos(x)}{\lim_{x\to\infty} 1+\sin(x)}$$ – Dave Nov 8 '17 at 19:01
• 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. – David Mitra Nov 8 '17 at 19:02
• The question has an answer here math.stackexchange.com/questions/2500171/… – Rene Schipperus Nov 8 '17 at 19:13

• $f'(x)/g'(x)$ 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).