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. 

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.
