Finding the limit $\lim_{x\to\infty}\left(1 + a^x\right)^{(1/x)}$ for $a > 0$

Question

Let $a > 0$. I am trying to determine rigorously the limit $\lim_{x\to\infty}\left(1 + a^x\right)^{(1/x)}$. If $0 < a < 1$, it is quite easy to see that the limit is 1. Similar is the case for $a = 1$ as then $\lim_{x\to\infty}(2)^{(1/x)} = 1$. So we may assume that $a > 1$. Empirically the limit seems to be $1 + a$, but I do not know how to put the pieces together. Namely, I am really, really, bad at root/exponential $\epsilon-\delta$ limit proofs and do not know how I even should choose the $\delta$. All hints and tips are appreciated!

Answer 1

This won't be rigorous but it's a start. As $x\to +\infty$ we have for $a > 1$

$$(1+a^x)^{1/x} = \sqrt[x]{1+a^x} \sim \sqrt[x]{a^x} \to a$$

The value of $a$, if $a> 1$, won't matter.

Consider that the limit $\lim_{x\to +\infty} \sqrt[x]{a^x}$ is a notable limit. Its generalisation to $\lim_{x\to +\infty} \sqrt[x]{a^x + k}$ for $k\in\mathbb{R}$ follows easily.

Answer 2

Note that for $a>1$,

$$ a<(1+a^x)^\frac{1}{x}<(2a^x)^\frac{1}{x}=a\cdot 2^\frac{1}{x} . $$

Answer 3

Take $\log (1 + a^x)^{ \frac 1 x}$ and see where the limit as $x \to \infty$ tends to (say, with L'Hôpital). Since the logarithm is continuous, this gives you your answer.