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
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.
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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}(1+3^n)^{1/n}=3}$$
