[tex]\text{Suntem in cazul de nedeterminare }\dfrac{\infty}{\infty}\text{ deci putem aplica lema Stolz-Cesaro.}\\\displaystyle \limit \lim_{n\to\infty} \dfrac{\ln n}{n}=\limit\lim_{n\to\infty}\dfrac{\ln(n+1)-\ln n}{n+1-n} =\limit\lim_{n\to\infty} \ln\left(\dfrac{n+1}{n}\right)=\ln 1=\boxed{0}[/tex]
