Showing that $\lim_{x\to 0} x\log{x}= 0$ without using L'Hopital's rule

Question

Is there a way of showing this limit without using L'Hopital's rule? Specifically using the substitution $u = -\log{x},$ with knowledge of the limit of $\lim_{x\to0+}\log{x}.$ Thank you.

Answer 1

If $u = -\log x$, then $e^{u} = e^{-\log x} = (e^{\log x})^{-1} = x^{-1} = 1/x$, that is, $x = \frac{1}{e^{u}}$.

If $x\to 0$, $\log x \to -\infty$. Thus,

$$\lim_{x \to 0}x\log x = \lim_{u \to \infty}-\frac{u}{e^{u}}$$

Answer 2

Using substitution $t=\frac{1}{x}$ gives: $x\log{x}=-\frac{\log{t}}{t}$ and $t\rightarrow \infty$. Now notice that for $t>2; 0<\log{t}<\sqrt{t}$ so: $$0<\frac{\log{t}}{t}<\frac{1}{\sqrt{t}}$$ Taking the limit when $t\rightarrow\infty$ gives the desired result.