Why does this limit work?

Question

Let $h(x)= (1+1/x)^x$ and $g(x)$ be another function.

Now suppose $\lim\limits_{x \to \infty} g(x)= \infty$. Then $\lim\limits_{x \to \infty} h(g(x))$ =$\lim\limits_{x \to \infty} h(x)=e$.

I would like to see a rigorous proof(or a reference for the proof) why this is true ? Exactly, what properties of limit are being used to get this result?

Answer 1

Write $\lim_{x \rightarrow \infty} h(x) = C \in \mathbb{R} \cup \{\pm \infty\}$ if and only if, for any sequence $(x_n)_{n \in \mathbb{N}}$ with $x_n \rightarrow \infty (n \rightarrow \infty)$, $\lim_{n \rightarrow \infty} h(x_n) = C$.

Assume you know that $\lim_{x \rightarrow \infty} h(x) = e$. If not, you can prove this by taking logarithms.

Now for any sequence $x_n \rightarrow \infty$, $g(x_n) \rightarrow \infty$ by assumption on $g$, so $\lim_{n \rightarrow \infty} h(g(x_n)) = \lim_{x \rightarrow \infty} h(x) = e$. Therefore $\lim_{x \rightarrow \infty} h(g(x))$ = e.

Answer 2

I assume you realize that $h$ is continuous and its limiting value at infinity is $e$.

Now as $h$ is continuous, you can pull the limit inside, so that

$$ \lim_{x \to \infty} h(g(x)) = h\left(\lim_{x \to \infty} g(x)\right) = \lim_{y\to\infty} h(y) = e$$

Added: Since $\lim\limits_{x\to\infty} h(x) = e$, for every $\epsilon > 0$, $\exists M > 0$ such that $|h(x) - e| < \epsilon$ for all $x \notin [-M,M]$.

Also, as $\lim\limits_{x\to\infty}g(x) = \infty$, $\exists N > 0$ such that $|g(x)| > M$ for all $x \notin [-N,N]$.

Hence we have for all $x \notin [-N,N]$,

$$ |h(g(x)) - e| < \epsilon$$.

whence $\lim\limits_{x\to\infty}h(g(x)) = e$.