What is the limit $\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}$? Can you try to solve it? I tried to do something but I do not know how to continue it:
\begin{align}
& \lim_{n\to\infty}\sqrt[n]{n^2+3n+1}=\lim_{n\to\infty}(n^2+3n+1)^{1/n} = e^{\lim_{n\to\infty} \frac{1}{n}\ln(n^2+3n+1)} \\[10pt]
= {} & e^{\lim_{n\to\infty}\frac{1}{n} \frac{\ln(n^2+3n+1)}{n^2+3n+1} (n^2+3n+1)}.
\end{align}

Hint: $$ \frac{1}{n} \ln \left( n^2+3n+1 \right) = \frac{1}{n} \ln \left( n \left(n+3+\frac{1}{n} \right) \right) = \frac{1}{n} \ln n + \frac{1}{n} \ln \left( n+3+\frac{1}{n} \right) $$
Write in the last expression: 
$$ \frac{1}{n} = \frac{1}{n}-\frac{1}{n+3+\frac{1}{n}}+\frac{1}{n+3+\frac{1}{n}}= \frac{1}{n^2} \frac{3n+1}{n+3n+1} +\frac{1}{n+3+\frac{1}{n}}$$
