Prove that $\lim\limits_{x \to 0^+}x^a \ln x =0$ for all $a >0$, without L'Hôpital

Question

Prove that, for every $a>0$,

$$\lim_{x \to 0^+}x^a \ln x =0$$

without L'Hôpital's rule.

My Try :

$$u=x^a \to \ln u=a\ln x \to x=\exp (\dfrac{\ln u}{a})$$

So we have :

$$\lim_{x \to 0^+}u \ln (\exp (\dfrac{\ln u}{a})) =\lim_{x \to 0^+}\dfrac{u}{a} \ln u =$$

now what ?

Answer 1

Write $$ L=\lim_{x \to 0^+}x^{a}\ln(x) = \lim_{x \to 0^+}\frac{\ln(x)}{x^{-a}}$$

and change the variable $\displaystyle{x=\frac{1}{u}}$, so $u\to \infty$

$$\underset{u\to\infty}{\lim}{\ \frac{\ln\left(\frac{1}{u}\right)}{u^{a}}}=\underset{u\to\infty}{\lim}{\ \frac{-\ln(u)}{u^{a}}}$$ then $u=e^{t}$, so $$L=\underset{t\to\infty}{\lim}{\ \frac{-t}{e^{at}}}=0$$

Because: If $p(t)$ is a polynomial and $a>0$, then $$\underset{t\to\infty}{\lim}{\ \frac{p(t)}{e^{at}}}=0$$

Recall that $x^{-a}=e^{-a\ln(x)}$ so you can also say that:

$$L=\lim_{x \to 0^+}\frac{\ln(x)}{e^{-a\ln(x)}}=\underset{v\to-\infty}{\lim}{\ \frac{v}{e^{-av}}}=0$$ and change the variable $x$ to $\ v=\ln(x)\to-\infty\ $ as $\ x\to 0$.

Answer 2

For $a\gt0$, we have $u=x^a\to0^+$ as $x\to0^+$, and therefore

$$\lim_{x\to0^+}x^a\ln x={1\over a}\lim_{x\to0^+}x^a\ln x^a={1\over a}\lim_{u\to0^+}u\ln u$$

so it remains to show that $\lim_{x\to0^+}x\ln x=0$ without resorting to L'Hopital. My favorite way of doing this is to use the integral definition of the natural logarithm, $\ln x=\int_1^x{dt\over t}$, and some crude inequalities based on the decreasing nature of $1/t$:

For $0\lt x\lt1$, we have

$$0\le x|\ln x|=x\int_x^1{dt\over t}=x\int_x^\sqrt x{dt\over t}+x\int_\sqrt x^1{dt\over t}\le x\left(\sqrt x-x\over x\right)+x\left(1-\sqrt x\over \sqrt x \right)\\=2(\sqrt x-x)\to0$$