Squeeze/Sandwich Theorem Involving $n^{th}$ root: $\lim _{ n\rightarrow \infty }{\left(3^n+1\right)}^{\frac 1n}$ Find $\lim _{ n\rightarrow \infty  }{ { \left( { 3 }^{ n }+1 \right)  }^{ \frac { 1 }{ n }  } } $ using the squeeze theorem
I have come across ways to do this but none mention the squeeze (or sandwich) theorem. I know I need to find $2$ functions which squeeze the given function but can only think of using $(3^n)^{1/n}$ i.e. $3$ as the $\le $ function
 A: As you pointed out in the question, $(3^n+1)^{\frac{1}{n}}\geq 3$, and on the other hand
$$ (3^n+1)^{\frac{1}{n}}\leq (3^n+3^n)^{\frac{1}{n}}=3\cdot 2^{\frac{1}{n}}$$
Since $\lim_{n\to\infty}2^{\frac{1}{n}}=1$, it follows that $\lim_{n\to\infty}(3^n+1)^{\frac{1}{n}}=3$ by the squeeze theorem.
A: 
PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities 
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{1+x}\le \log(1+x)\le x} \tag 1$$
for $x>-1$.

Note that we can write
$$\begin{align}
(1+3^n)^{1/n}&=3\left(1+\frac1{3^n}\right)^{1/n}\\\\
&=3e^{\frac1n \log\left(1+\frac{1}{3^n}\right)}\tag 2
\end{align}$$
Applying $(1)$ to $(2)$ we find that
$$\begin{align}
3e^{\frac{1}{n(1+3^n)}}\le (1+3^n)^{1/n}\le 3e^{\frac{1}{n3^n}} \tag 3
\end{align}$$
Finally, applying the squeeze theorem to $(3)$ reveals the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}(1+3^n)^{1/n}=3}$$
