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

Question

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 ??

The numerator and denominator have to be differentiable around the point of interest; not necessarily at it. — parsiad, Aug 17 '19 at 09:20
@parsiad Neighborhood of infinity is infinity, in that case too, it shouldn't hold — Mr.HiggsBoson, Aug 17 '19 at 09:21
Make the change of variables $y = 1/x$. Now the limit argument is zero and you can look at a neighbourhood of zero. There are some subtleties with the definition of log here but you can make it work. — parsiad, Aug 17 '19 at 09:23
@parsiad When is a change of variable allowed. Can I always make a substitution ?? — Mr.HiggsBoson, Aug 17 '19 at 09:24
@RandomAspirant you are fine using L'Hopitals here - the derivatives are well defined. — Blueyedaisy, Aug 17 '19 at 09:31
Neighborhood of infinity is an interval of the form $(a, \infty) $ and the functions involved here are indeed differentiable on $(0,\infty) $. All the other hypotheses of L'Hospital's Rule are also met here and hence its use is perfectly fine here. — Paramanand Singh, Nov 15 '19 at 02:02

Robert Tumasian III · Answer 1 · 2019-11-14T04:17:15.307

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.

score 0 · Answer 2 · answered Jan 03 '22 at 18:44

0

let k a positive real number ,

you could consider the following subsitution $ t=x^k$ , in fact : $\lim_{x\to\infty} \frac{ln(x)}{x^k} $= $\lim_{t\to\infty} \frac{ln(t)}{t*k} $

answered Jan 03 '22 at 18:44

Oussama Akar

339

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

2 Answers2