6

It's well known that $\displaystyle{\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^x = e^k}$.

Suppose $xf(x) \to k$ as $x \to \infty$. Do we necessarily have $(1 + f(x))^x \to e^k$?

If $f$ happens to be differentiable, and if we also have $\displaystyle{\frac{f'(x)}{-x^{-2}} \to k}$, then the answer to the above question is yes, as can be seen by an application of L'Hopital's rule. But without assuming this second limit exists, I don't think L'Hopital's rule is relevant.

If it's any easier, I'd also be interested in the answer for the special case when $k = 1$.

This is a problem I thought of as I was solving problem 3.2.1(b) (page 72) from "Problems in Mathematical Analysis I" by Kaczor and Nowak.

Andrew Kelley
  • 1,798

1 Answers1

7

If $\lim_{x\to\infty}xf(x)=k$, then $\lim_{x\to\infty}f(x)=0$ and $f(x)=\frac{k+o(1)}{x}$.

METHODOLOGY $1$:

We assume in the following that $k>0$ although analogous analysis applies for $k< 0$. Note that we can write

$$\begin{align} \left(1+f(x)\right)^x&=\left(1+\frac kx +o\left(\frac{1}{x}\right)\right)^x\\\\ &=\left(1+\frac kx \right)^x\left(1+\frac {o\left(\frac{1}{x}\right)}{1+\frac kx}\right)^x\tag1 \end{align}$$

Now, applying Bernoulli's Inequality reveals (for $x$ sufficiently large)

$$\begin{align} \left(1+\frac {o\left(\frac{1}{x}\right)}{1+\frac kx}\right)^x\ge 1+\frac{o\left(1\right)}{1+\frac kx}\tag2 \end{align}$$

In addition, we have from Bernoulli's Inequality (for $x$ sufficiently large)

$$\begin{align} \left(1+\frac {o\left(\frac{1}{x}\right)}{1+\frac kx}\right)^x\le \frac1{\left(1-\frac {o\left(\frac{1}{x}\right)}{1+\frac kx}\right)^x}\le \frac1{1-\frac{o\left(1\right)}{1+\frac kx}}\tag3 \end{align}$$

Putting together $(2)$ and $(3)$ and applying the squeeze theorem we find that

$$\lim_{x\to\infty}\left(1+\frac {o\left(\frac{1}{x}\right)}{1+\frac kx}\right)^x=1\tag4$$

Finally, using $(4)$ in $(1)$ yields the coveted limit

$$\lim_{x\to\infty}\left(1+f(x)\right)^x=e^k$$

METHODOLOGY $2$:

We begin by writing

$$\begin{align} \left(1+f(x)\right)^x&=e^{x\log(1+f(x))}\\\\ &=e^{x\left(f(x)+O\left(f(x)^2\right)\right)}\tag5 \end{align}$$

Letting $x\to \infty$ in $(5)$, we find that

$$\lim_{x\to\infty}\left(1+f(x)\right)^x=e^k$$

as was to be shown!

Alternatively, since we can write

$$\frac{f(x)}{1+f(x)}\le \log(1+f(x))\le f(x)$$

Then we have the inequalities

$$e^{xf(x)/(1+f(x))}\le \left(1+f(x)\right)^x\le e^{xf(x)}$$

whence application of the squeeze theorem yields

$$\lim_{x\to\infty}\left(1+f(x)\right)^x=e^k$$

Mark Viola
  • 179,405
  • I had the second method in mind as well. Great answer as always! – Cameron Williams Jul 12 '21 at 03:07
  • @CameronWilliams Hi Cameron! Thank you! And how are you? It's been a while. – Mark Viola Jul 12 '21 at 03:12
  • I've been well. How about you? – Cameron Williams Jul 12 '21 at 12:02
  • Method 2 makes sense (including the "alternatively" part). I also see how to change Method 1 to get another proof, but as stated, I believe (2) incorrectly applies Bernoulli's inequality. Namely, replacing $o(1/x)$ by $-1/x^2$ (and letting $k = 1$) gives a counterexample to inequality (2), as checked by wolframalpha. – Andrew Kelley Jul 23 '21 at 02:04
  • @AndrewKelley Bernoulli's inequality states that $(1+t)^s\ge 1+st$ for every $s\ge 0$ and every $t>-1$. Here $x\ge0$ and certainly for $k>0$ and $x$ sufficiently large, $$o(1/x)/(1+k/x) >-1$. So, everything works. – Mark Viola Jul 23 '21 at 02:25
  • @Mark, yes, I understand the statement of Bernoulli's inequality, but I think you made an algebra mistake. In particular, when applied to $(1 + o(1/x)/(1+k/x))^x$, Bernoulli's inequality gives a lower bound of $1 + x * o(1/x)/(1 + k/x) = 1 + x^2 * o(1/x)/(x+k)$ – Andrew Kelley Jul 23 '21 at 02:33
  • I had a typo in my comment from 2 minutes ago (forgetting the divide by symbol in the last expression) – Andrew Kelley Jul 23 '21 at 02:36
  • The problem is that if $o(1/x)$ is negative, then $1 + x^2 * o(1/x)/(x+k)$ is not larger than $1 + o(1/x)/(x+k)$. The issue is not whether or not Bernoulli's inequality applies. It does. The issue is your lower bound does not match what Bernoulli's inequality gives. – Andrew Kelley Jul 23 '21 at 02:38
  • 1
    @AndrewKelley Ah. The typographical errata have been edited. Thank you for the catch! – Mark Viola Jul 23 '21 at 02:44
  • No problem. Thank you for your thorough answer. I appreciated that you gave multiple approaches. (It did take me some effort to fill in the details though.) – Andrew Kelley Jul 23 '21 at 02:51
  • You're welcome. My pleasure. – Mark Viola Jul 23 '21 at 02:57