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}
 A: For positive values of $n$ we have$$\sqrt[n]{n^2}\lt \sqrt[n]{n^2+3n+1}\le\sqrt[n]{5n^2}$$ Note that $$\lim_{n\to\infty}\sqrt[n]{n}=\lim_{n\to\infty}\sqrt[n]{5}=\color{Red}{1}.$$ Hence by the squeeze we have the answer.
A: $$
\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\sim \lim_{n\to\infty}\sqrt[n]{n^2}=\lim_{n\rightarrow \infty}n^{2/n}\rightarrow1
$$
If this is not satisfying, you can also note that
$$
(n+2)^2=n^2+4n+2
$$
Which is always larger than $n^2+3n+1$ and you can squeeze
$$
\lim_{n\to\infty}\sqrt[n]{n^2}\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq\lim_{n\to\infty}\sqrt[n]{(n+2)^2}\\
\Rightarrow
\lim_{n\to\infty}n^{2/n}\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq\lim_{n\to\infty}{(n+2)^{2/n}}\\
\Rightarrow 1\leq\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}\leq 1
$$
A: 
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!
A: 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: 
$$ \frac{1}{n} = \frac{1}{n}-\frac{1}{n+3+\frac{1}{n}}+\frac{1}{n+3+\frac{1}{n}}= \frac{1}{n^2} \frac{3n+1}{n+3n+1} +\frac{1}{n+3+\frac{1}{n}}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{n \to \infty}{1 \over n}\ln\pars{n^{2} + 3n + 1}
\\[5px] = &\
\lim_{n \to \infty}
\ln\pars{\bracks{n + 1}^{\, 2} + 3\bracks{n + 1} + 1 \over n^{2} + 3n + 1}\qquad
\pars{~Stolz-Ces\grave{a}ro\ Theorem~}
\\[5mm] = &\ \color{#f00}{0} \qquad\implies\qquad
\lim_{n \to \infty}\root[n]{n^{2} + 3n + 1} = \exp\pars{\color{#f00}{0}} = \bbx{\ds{1}}
\end{align}
