The value of $(2^n+3^n+4^n)^{1/n}$ as $n \rightarrow \infty?$ I was thinking about the following problem:
How can i find the value of $(2^n+3^n+4^n)^{1/n}$ as $n \rightarrow \infty?$
Can someone point me in the right direction? Thanks in advance for your time.

Here you have an approach that uses the definition of the constant $e=\lim_{n\to\infty}(1+\frac 1n)^n$.
$$
  (2^n+3^n+4^n)^{\frac 1n}
~ = ~~
  4
  \left(
    1 + \frac{2^n+3^n}{4^n}
  \right)^{n\frac{2^n+3^n}{4^n}\frac{4^n}{2^n+3^n}}
$$
Since $n\frac{2^n+3^n}{4^n}\to 0$ and $(1 + \frac{2^n+3^n}{4^n})^{\frac{4^n}{2^n+3^n}}\to e$ you have that the above limit is equal to $4e^0=4$.
