Squeeze Theorem for $(2^n + 3^n + 4^n)^{\frac{1}{n}}$ How to find this limit when $n \rightarrow \infty$.
$$(2^n + 3^n + 4^n)^{\frac{1}{n}}.$$
My Idea:
We need to squeeze this sequence.
$$ (4^n)^{\frac{1}{n}}\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le (3\cdot4^n)^{\frac{1}{n}}$$
So that we obtain
$$ 4\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le (3)^{\frac{1}{n}} 4$$
By taking the limit $n \rightarrow \infty$
$$4\le (2^n + 3^n + 4^n)^{\frac{1}{n}} \le 4 $$
So, by squeeze theorem
$$\lim_{n \to \infty} (2^n + 3^n + 4^n)^{\frac{1}{n}} = 4.$$
Actually, I still doubt about my idea.
Is my idea correct? Thanks in advance.

I have another idea that might be useful:
$$
(2^n+3^n+4^n)^\frac{1}{n}=\frac{4}{4}(2^n+3^n+4^n)^\frac{1}{n}=4\left(\frac{2^n+3^n+4^n}{4^n}\right)^\frac{1}{n}=4\left(\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n+1\right)^\frac{1}{n}
$$
Taking limits you get $4$
