Evaluate the limit $\lim_{n \to \infty} \left(3^n+1\right)^{\frac1n} $ How do you evaluate this sequence limit using the squeeze/sandwich theorem?
$$\lim_{n \to \infty} \left(3^n+1\right)^{\frac1n} $$
I don't really know where to start. I've tried using the fact that $\lim_{n \to \infty} \left(3^n\right)^{\frac1n} = 3$ (which is the correct answer) but I don't know where to go from there.
Thanks!
 A: We have that
$$3=(3^n)^{1/n}\le (3^n+1)^{1/n}\le (3^n+3^n)^{1/n}=3\cdot2^\frac1n$$
then conclude by squeeze theorem.
A: You can use  $3=(3^n)^{1/n} \leq(3^n + 1)^{1/n} \leq (3^n+3^n)^{1/n}=2^{1/n}\cdot 3$
A: A slightly different way is taking $3^n$ out of $(3^n+1)^{1/n}$, that is
$$
(3^n+1)^{1/n}=3(1+3^{-n})^{1/n}
$$
Now note that $1\leqslant 1+3^{-n}\leqslant 2$ for every $n\in \mathbb N $, therefore taking limits in the inequality we arrive at
$$
1\leqslant \lim_{n\to\infty}(1+3^{-n})^{1/n}\leqslant \lim_{n\to\infty}2^{1/n}=1
$$
and so
$$
\lim_{n\to\infty}(3^n+1)^{1/n}=\lim_{n\to\infty}3(1+3^{-n})^{1/n}=3
$$
A: Consider $y = (3^n + 1)^\frac{1}{n}$ now affect logarithm to both sides:$$\ln{y} = \frac{\ln{(3^n + 1)}}{n}$$ obviouslly if $n$ goes to infinity we can omit 1 inside the logarithm then we easily obtain: $\ln{y} = \ln 3$ when $n$ goes to infinity.
so the answer is: $$y = 3$$
A: With logarithm: rewrite the expression as
$$
e^{\frac{1}{n}(\log 3^n + \log (1+\frac{1}{3^n})}
$$
The first term is $3$. The second has easy bounds:
$$
0<\log (1+\frac{1}{3^n})<\log 2
$$
and, therefore,
$$
1<e^{\frac{1}{n}\log (1+\frac{1}{3^n})}<e^{\frac{\log 2}{n}} \to_n 1
$$
A: $$
\displaystyle\left(3^{n} + 1\right)^{1/n} =
3\left(1 + {1 \over 3^{n}}\right)^{1/n} =
3\left[\left(1 + {1 \over 3^{n}}\right)^{3^{\large n}}\right]^{1/\left(3^{\large n}n\right)}
\,\,\,\stackrel{\mathrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\bbox[#ffd,10px,border:1px groove navy]{\large 3}
$$
A: Where $n$ is sufficiently large $3^n$ is much greater that $1$, which can be neglected (we can notice that $100000000000000000000$ and $100000000000000000001$ are "almost" the same).
So $3^n+1 \sim_{n \to \infty} 3^n$ by the fact that $\lim_{n \to \infty} \frac{3^n+1}{3^n}=1$ fastly and the rest can be done easily.
