4

I'm familiar with the result is that for $a \in \mathbb{R}$: $$ \lim_{n \to \infty} \left(1+\frac{a}{n} \right)^n = e^a $$ I'm just wondering, if given something like $\lim_{n\to \infty}f(n) =d$, then the general result should be:

$$ \lim_{n \to \infty} \left(1+\frac{f(n)}{n} \right)^n = e^{\lim_{n \to \infty} f(n)}=e^d $$

Cameron Williams
  • 29,432
WeakLearner
  • 5,982

2 Answers2

6

Yes, the result is true. When $n$ goes to infinity, since $\frac{f(n)}{n}$ goes to $0$, we may write

$$\begin{eqnarray*}\left(1 + \frac{f(n)}{n}\right)^n &=& \exp \left[n \ln\left(1 + \frac{f(n)}{n}\right)\right]\\ &=& \exp \left[n \ln\left(1 + \frac{d + o(1)}{n}\right)\right]\\ &=& \exp \left[n \left( \frac{d}{n} + o\left(\frac{1}{n} \right) \right)\right]\\ &=& \exp \left[d + o(1)\right]\\ &=& e^d + o(1) \underset{n \to \infty}{\longrightarrow} e^d \end{eqnarray*}$$

Joel Cohen
  • 9,289
  • I was writing quite a similar argument to yours but without the $\exp$. Since it was basically the same, I deleted mine. $+1$ – Cameron Williams Aug 01 '15 at 02:38
  • @JoelCohen Thanks, that makes sense, Is it incorrect to write the limit in the exponent as I have it? – WeakLearner Aug 01 '15 at 02:49
  • 1
    @dimebucker91 : You're welcome :) I think what you wrote is correct, indeed $\lim_{n \to \infty} f(n)$, however complicated, is just a number after all. – Joel Cohen Aug 01 '15 at 02:51
3

Suppose $f(n) \to d$ as $n \to \infty$, and let $\epsilon > 0$ be arbitrary. Then, for $N$ sufficiently large (say $n > N$), we have

$$\left(1+\frac{d-\epsilon}{n}\right)^n \le \left(1+\frac{f(n)}{n}\right)^n \le \left(1+\frac{d+\epsilon}{n}\right)^n$$

if we take $n$ large enough (say, $n>M$), then we can also conclude

$$e^{d-\epsilon}-\epsilon \le \left(1+\frac{f(n)}{n}\right)^n \le e^{d+\epsilon}+\epsilon$$

But, this is true for all $\epsilon > 0$ provided $n$ is sufficiently large, so we can conclude

$$\lim_{\epsilon\to 0} e^{d-\epsilon}-\epsilon \le \liminf_{n \to\infty} \left(1+\frac{f(n)}{n}\right)^n$$ and also $$\limsup_{n \to\infty} \left(1+\frac{f(n)}{n}\right)^n \le \lim_{\epsilon \to 0} e^{d+\epsilon}+\epsilon$$

from which we have $\liminf_{n \to\infty} \left(1+\frac{f(n)}{n}\right)^n = \limsup_{n \to\infty} \left(1+\frac{f(n)}{n}\right)^n = e^d$, so $\lim_{n \to\infty} \left(1+\frac{f(n)}{n}\right)^n=e^d$