TITLE: The Limit: $\lim_{x \to \infty}\frac{e^{f(x+a)}}{e^{f(x)}}$
QUESTION [6 upvotes]: I'm doing some challenge review problems and I was wondering whether this proof looked correct:
Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function with $\lim_{x \to \infty}f'(x)=1$ and $a \in \mathbb{R}$. Prove the limit exists and find it.
$$
\lim_{x \to \infty} \frac{e^{f(x+a)}}{e^{f(x)}}
$$
Here is what I did,
$$
\begin{align}
\lim_{x \to \infty} \frac{e^{f(x+a)}}{e^{f(x)}}&=\lim_{n \to \infty}e^{f(x+a)}e^{-f(x)} \\
&=\lim_{x \to \infty}e^{f(x+a)-f(x)} \\
&=\lim_{x \to \infty}\left(e^{f(x+a)-f(x)}\right)^\frac{a}{a} \\
&=\lim_{x \to \infty}\left(e^\frac{f(x+a)-f(x)}{a}\right)^a \\
&=\left(e^{\lim_{x \to \infty}\frac{f(x+a)-f(x)}{a}}\right)^a \\
&=\left(e^{\lim_{x \to \infty}f'(x)}\right)^a \\
&=(e^1)^a\\
&=e^a
\end{align}
$$
Though missing verbal explanation of the individual steps to be a 'good' proof, does this look like the right idea?

REPLY [4 votes]: Your proof requires some additional justification since
$$ \frac{f(x+a)-f(x)}{a}\ne f'(x)$$
in general. However, note that since $L=\lim\limits_{x\to \infty}f'(x)$ exists, for any $\epsilon >0$ we have some $N$ such that $x\ge N\implies |f'(x)-L|<\epsilon$, so for $x\ge n$ we have
$$\begin{align}
\left|\frac{f(x+a)-f(x)}{a}-L\right|
&=\left|\frac{1}{a}\int_x^{x+a}(f'(t)-L)dt\right|\\
&\le \frac{1}{a}\int_{x}^{x+a}|f'(t)-L|dt\\
&\le \frac1a\int_x^{x+a}\epsilon = \epsilon
\end{align}$$
thus $\lim\limits_{x\to\infty}\frac{f(x+a)-f(x)}{a}=L$