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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.

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    You can do l'Hospital as many times as wanted/required as long as the conditions are fullfiled. – 2017-01-02
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    Yes, you can use l'hopital more than once. – 2017-01-02
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    @DonAntonio thanks ^_^ – 2017-01-02
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    @amWhy yeah, my bad – 2017-01-02
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    Why exactly was this put on hold? The question has both context and appropriate details. – 2017-01-03
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    @Ennar I have no clue :( – 2017-01-03

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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: $$\lim_{x \to \infty} \frac{x}{\ln^2 x} = \lim_{x \to \infty} \frac x2 = \infty$$

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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: $\lim\frac{f(x)}{g(x)} = L$.

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    Why was this downvoted? – 2017-01-03