$\lim_{n\rightarrow \infty}(3^n+7^n)^{1/n}$

Question

I was trying to solve this problem about limit and I have some problems. $$\lim_{n\rightarrow \infty}(3^n+7^n)^{1/n}$$ I need some help with this limit please.

Answer 1

HINT: We have $$7^n < 3^n + 7^n < 2\cdot 7^n$$

Answer 2

$$ (3^n+7^n)^{\frac1n}=\left[7^n\left(\frac{3^n}{7^n}+1\right)\right]^{\frac{1}{n}}=7(1+q^n)^\frac{1}{n} \quad \forall n\in \mathbb{N} $$ with $q=\frac37<1$. Taking the limit we get $$ \lim_n(3^n+7^n)^{\frac1n}=\lim_n7(1+q^n)^\frac{1}{n}=7. $$

Answer 3

The sandwich lemma states that for any sequence $(a_n)$ with $(b_n)$, $(c_n)$ such that $b_n < a_n < c_n$ for any given natural number $n$, then for any $x$:

$$b_n \to x \text{ and } c_n \to x \implies a_n \to x$$

If for any given $n$, $a_n$ is $(3^n + 7^n)^{1/n}$, you need to find suitable sequences for $(b_n)$ and $(c_n)$ with the same limit as $n \to \infty$ which "sandwich" $(a_n)$.

Answer 4

$$\lim_{n\rightarrow \infty}(3^n+7^n)^{1/n}=7\lim_{n\rightarrow \infty}\left(\left(\frac37\right)^n+1\right)^{1/n}.$$

The limit on the right is $1$ as

$$1<\left(\left(\frac37\right)^n+1\right)^{1/n}<\left(\frac37\right)^n+1.$$