The Limits of L’Hôpital

This may be checked directly: $\lim_{x\rightarrow\infty} \frac{x + \sin x}{x + 1} = \lim_{x\rightarrow\infty} \frac{1 + \tfrac{\sin x}{x}}{1 + \tfrac{1}{x}} = \frac{1 + 0}{1 + 0} = 1.$ $\lim_{x\rightarrow\infty} \frac{\tfrac{d}{dx}(x + \sin x)}{\tfrac{d}{dx}(x + 1)} = \lim_{x\rightarrow\infty} \frac{1 + \cos x}{1} = 1 + \lim_{x\rightarrow\infty}\cos x\mbox{, which does not exist!}$ Obviously, the two sides aren't equal!

So why doesn't L’Hôpital's rule work in this limit? We can easily check that the original limit had the indeterminate limit form $\frac{\infty}{\infty}$ , after all. The problem, it turns out, is that most people either have forgotten or never learned the complete statement of L’Hôpital's rule, so here it is, in its entirety:

L’Hôpital's Rule : If $\displaystyle\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$ is an indeterminate limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ AND $\displaystyle \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$ exists [i.e. it either equals a real number or $\pm\infty$ ], then $\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}.$

The moral of this story is that you can still use L’Hôpital's rule as usual, but if you end up with a limit that doesn't exist, then you can't claim the original limit doesn't exist.

We can evaluate the limit like this $\lim_{x\rightarrow\infty} \frac{x + \sin x}{x + 1} = \lim_{x\rightarrow\infty} \frac{1 + \tfrac{\sin x}{x}}{1 + \tfrac{1}{x}} = \frac{1 + 0}{1 + 0} = 1.$

So No $\dfrac{0}{0}$ form.

Hence So L'Hospital cannot be applied.

Moral of the question- Dont use theorems without knowing its reason of coming and make your base stronger.

No matter how large the value of x is, sine and cosine will just throw value in [-1, 1] range. No need for L'Hospital' s.

49% got right. everyone probably just guessed.