TITLE: Limit question - L'Hopital's rule doesn't seem to work
QUESTION [26 upvotes]: I have been recently trying to solve this limit problem. First of all, I used L'Hopital's rule but it doesn't seem to work (because I thought that this limit is of form $\frac{\infty}{\infty}$). Am I doing it correctly? I don't seem to understand where am I wrong.
$$\lim_{x \to \infty} \left(\frac{x+\sin^3x}{5x+6}\right)$$

REPLY [9 votes]: One of the conditions of applying L'Hospital's Rule is that $f'(x)/g'(x)$ must exist.
$$\lim\limits_{x \to \infty} \frac{f'(x)}{g'(x)}$$
After one application of L'Hospitals, you arrived at a finite numerator over a finite denominator.  But while the numerator was finite, it was non-convergent, and so that limit did not exist.  Just like the much simpler $\sin(x)$ does not converge to a single value when x approaches infinity -- it oscillates between +/-1. $$\lim\limits_{x \to \infty} {\sin(x)}$$
So all of L'Hospital's pre-conditions must exist for you to use it.  As others have mentioned, this limit could more easily be solved by using the squeezing theorem.  The numerator's value gets squeezed between $x+1$ and $x-1$. Both of those limits go to $1/5$.