Evaluate $\lim\limits_{x\to0} x^p (\log(1/x))^m$ where $00$, or $p>1 $ and $m>0$

Question

how to find the limit of the following:

$\lim\limits_{x\to0} x^p (\log(1/x))^m$

where

(1) $0<p<1$ and $m>0$

(2) $p>1$ and $m>0$

explanation: In case (1)and (2) both, it is of $(\frac{0}{0})$ form, hence by using L-Hospital Rule:

Let $T = \lim\limits_{x\to0}\frac {x^p} {(\log(1/x))^{-m}}$

$~~~~~~~~~ = \lim\limits_{x\to0}\frac {px^p}{m (\log(1/x))^{-(m+1)}}$

$~~~~~~~~~=~$T$~\lim\limits_{x\to0}\frac {p} {m}(\log(1/x))$

$=>T(1-\lim\limits_{x\to0}\frac {p} {m}(\log(1/x)))=0$

from this $~T$ should be $0.$

Whether the given solution is correct or is there any other method to solve this problem, please help me.

@Piquito that will be a different quetion and it is easy to get limit as zero but I need limit when $m>0$ — nazim khan, Oct 19 '16 at 15:58
@nazimkhan: in this case, with $m\gt 0$, you don't have the form $\frac 00$ but $0\cdot\infty$ — Piquito, Oct 20 '16 at 13:56

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 <x\tag 1$$

for $x>0$.

Using $(1)$ along with the identity $\log(x^a)=a\log(x)$ reveals

$$\log(1/x)< \frac{1}{ax^a} \tag 2$$

for $a>0$.

From $(2)$, we have for all $a>0$ and $x\le 1$

$$0\le x^p\log^m(x)< \frac1{a^m} x^{p-am} \tag 3$$

Since $(3)$ is true for all $a$, it is certainly true for $a<p/m$. In that case, application of the squeeze theorem yields the coveted limit

$$\lim_{x\to 0^+}x^p\log^m(x)=0$$

for all $p>0$ and $m>0$.

Evaluate $\lim\limits_{x\to0} x^p (\log(1/x))^m$ where $00$, or $p>1 $ and $m>0$

1 Answers1

Linked