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}


PRIMER
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x) \le x-1 \tag 1$$
for $x>0$.


We will now use $(1)$ to show that $\lim_{n\to \infty}\frac1n \log(n)=0$.  To do so, we see from $(1)$ that for any $\alpha>0$ 
$$\begin{align}
\alpha \log(x) &=\log(x^\alpha)\\\\
&\le x^\alpha -1\\\\
&<x^\alpha\\\\
\log(x)\le \frac{x^\alpha}{\alpha}
\end{align}$$
Hence, we can assert that 
$$\frac1n \log(n)\le \frac{n^{\alpha -1}}{\alpha} \tag 2$$
for any $\alpha >0$.  
Since $(2)$ is true for any positive $\alpha$, it is certainly true for $0<\alpha <1$.  Therefore, for $0<\alpha <1$ we see that $\lim_{n\to \infty}\frac1n \log(n)=0$.  

Next, using $\log(n^2+3n+1)\le \log(5n^2)=\log(5)+2\log(n)$, we see that
$$\begin{align}
\lim_{n\to \infty}\sqrt[n]{n^2+3n+1}&=\lim_{n\to \infty}e^{\frac1n \log(n^2+3n+1)}\\\\
&=e^{\lim_{n\to \infty}\frac1n \log(n^2+3n+1)}\\\\
&=e^0\\\\
&=1
\end{align}$$
as was to be shown!
