Evaluate the limit $\displaystyle\lim_{x \to 0^{+}} x \cdot \ln(x)$

I have this assignment

Evaluate the limit:

5.$\displaystyle\lim_{x \to 0^{+}} x \cdot \ln(x)$

I don't think we are allowed to use L'Hopital, but I can't imagine how else.

Solution 1:

If you set $y=\ln x$, you can rewrite it as $\lim_{x\to-\infty}ye^y=-\lim_{y\to\infty}\frac{y}{e^y}$, and you probably know what that is. Of course that approach requires that you be allowed to use known limits.

Solution 2:

Assume $0<x<1$, then $1/x>1$. Since $0\le\ln (1+t)\le t$ if $t>1$, we get $$0\le \frac{1}{2}\ln\frac{1}{x}= \ln\frac{1}{\sqrt{x}}=\ln\left(1+\frac{1}{\sqrt{x}}-1\right)\le \frac{1}{\sqrt{x}}-1$$ if $0<x<1$. Therefore $$0\le x\ln\frac{1}{x}\le 2(\sqrt{x}-x).$$ and we know that $\ln(1/x)=-\ln x$ holds.