Is applying L'Hopital's rule to $\lim_{x\to \infty}\frac{\ln x}{x ^k}, k \in R^+ $, because numerator and denominator isn't differentiable

Say I have a function $$\lim_{x\to \infty}\frac{\ln x}{x ^k}, k \in R^+ $$

So now the solution provided to me applied L'Hopital's Rule. But I'm unsure if it's correct or not.

Because for application of L'Hopital's, both numerator and denominator should be differentiable at the point where we are evaluating the limit.

But in this case, differentiating both $\ln x$ and $x^k$ at $\infty $ would yield that the derivative is $\to \infty$ . And so we should say it's not differentiable.

Also how can I solve/verify it without using L'Hopital's Rule and is the solution wrong or is there a problem in my understanding ??

Solution 1:

We can still apply L'hopital's Rule if the functions in the numerator and denominator are not differentiable at some $x$:

$\lim_{x\to\infty} \frac{ln(x)}{x^k} = \lim_{x\to\infty} \frac{\frac{d}{dx}ln(x)}{\frac{d}{dx}x^k} = \lim_{x\to\infty} \frac{\frac{1}{x}}{k\cdot x^{k-1}} = \lim_{x\to\infty} \frac{1}{x \cdot k \cdot x^{k-1}} $

$= \lim_{x\to\infty} \frac{1}{k\cdot x^k} = 0$ for some $k \in R^+$.

L'hopital's Rule is not necessarily needed, since the function in the denominator will grow more quickly than the function in the numerator as $x$ goes to $\infty$, and thus the limit is 0.