Computing limit - horizonal asymptote I tried to compute 
$$ \lim_{x \to \infty} \frac{x}{\ln^2 x},$$
but after using l'Hôpital's rule once I've got the expression 
$$ \frac{x}{2 \ln x}.$$
Is it OK to apply l'Hôpital's rule again, and conclude that
$$\lim_{x \to \infty} \frac{x}{2} = \infty \; ?$$
Thanks in advance.
 A: Yes, as @DonAntonio answered in the comments, you can use l'Hospital as many times as needed, provided that the preceding application of l'Hospital still renders a limit of indeterminate form. 
Assuming so, you proceeded quite nicely, using two applications of l'Hospital, and have correctly found the limit you sought:  $$\lim_{x \to \infty} \frac{x}{\ln^2 x} = \lim_{x \to \infty} \frac x2 = \infty$$
A: Let me try to explain why it is ok to apply L'Hospital as many times as needed. The theorem is of the following form:

If certain stuff is true, then: $$\exists\ \lim\frac{f'(x)}{g'(x)}
 \implies \left(\exists\ \lim\frac{f(x)}{g(x)}\ \wedge\
 \lim\frac{f(x)}{g(x)} = \lim\frac{f'(x)}{g'(x)}\right)$$

Now, let's say that we tried to calculate $\lim\frac{f^{(k)}(x)}{g^{(k)}(x)}$ for $k = 0,1,\ldots,n-1$ and every time we got indeterminate form (i.e. certain stuff were true for all $k=0,1,\ldots,n-1$) so we apply L'Hospital again to finally get $\lim\frac{f^{(n)}(x)}{g^{(n)}(x)} = L$.
Applying the above theorem first time we conclude that $\lim\frac{f^{(n-1)}(x)}{g^{(n-1)}(x)} = L$, second time: $\lim\frac{f^{(n-2)}(x)}{g^{(n-2)}(x)} = L$, ..., and $n$-th time: $\lim\frac{f(x)}{g(x)} = L$.
