Soluzione Esercizio 4

 

            \begin{align}\lim_{x \to 0^+} x\ln(x)&=\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}=\left[\frac{-\infty}{\infty}\right]\\ &=\xrightarrow{l'H \hat{o} pital} \lim_{x \to 0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x \to 0^+} -x=0\\ \end{align}