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?

  • $\begingroup$ Yep, you have the right idea. $\endgroup$ – David H Apr 12 '14 at 1:47
  • $\begingroup$ Except you have switched to $n \to \infty$ at one point, but otherwise this is good $\endgroup$ – user139388 Apr 12 '14 at 1:48
  • $\begingroup$ Looks good to me except for what I point out in my answer. I've edited your question to use $x$ consistently throughout; hopefully this is what you intended. $\endgroup$ – Alex Becker Apr 12 '14 at 2:01
  • $\begingroup$ @AlexBecker & user139388. Indeed, I meant $x$ for the limit throughout. I'm just so adjusted to $n$ I made the mental switch without even realizing it. Thanks for pointing it out! $\endgroup$ – Kyle L Apr 12 '14 at 2:05
  • $\begingroup$ @KyleL No problem. A latex tip for the future: using \left and \right around braces will automatically make them the correct size, no matter what is between them. $\endgroup$ – Alex Becker Apr 12 '14 at 2:06

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 $\lim\limits_{x\to\infty}\frac{f(x+a)-f(x)}{a}=L$

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  • $\begingroup$ Thank you! Indeed, I knew it required additional justification but couldn't immediately come up with a good argument but it seemed like that was the way the problem had to progress. $\endgroup$ – Kyle L Apr 12 '14 at 2:06
  • $\begingroup$ You should also be able to use the MVT here, since there's some $x_0$ in the vicinity of $x$ (specifically, in $[x, x+a]$) with $\frac{f(x+a)-f(x)}{a} = f'(x_0)$. $\endgroup$ – Steven Stadnicki Apr 12 '14 at 2:41

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