One of the standard definitions of $e$ is as

$$\lim_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^n$$

But in all cases I've seen this limit, it is proven as a limit of the sequence $\Big\{\big(1 + \frac{1}{n}\big)^n\Big\}$, which seems to cover the limit for only $n$ as an integer. Now my question is whether this sequence limit is equivalent to a normal limit for which $n$ can be any real number. I can think of cases which this isn't generally true;

$$\lim_{n\rightarrow\infty}\ \sin(n\pi)$$

comes readily to mind, for which the limit as a sequence is simply $0$ but as a general limit, it is undefined. The limit for $e$ is used exactly as if it were a normal limit, which leads me to believe it is equivalent. Are there conditions for which the limit of a sequence and the corresponding function are identical?

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    $\begingroup$ If $\lim_x f(x)$ exists, then $\lim_x f(x) = \lim_n f(n)$ (but as you point, there are cases where $\lim_n f(n)$ exists but $\lim_x f(x)$ does not). $\endgroup$ – Joel Cohen May 13 '12 at 19:32

If $x>0$ be a real number, then write $x=n+y_x$ where $0 \leq y_x \leq 1$.

It is easy then to show that

$$ \left(1+\frac{1}{n+1} \right)^n \leq \left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{n}\right)^{n+1} \,.$$

Using this, you can prove that if

$$e=\lim_n \left(1+\frac{1}{n}\right)^n$$ then $$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =e$$


To complete the answer, in general if $f(x)$ is monotonic, then for sure

$$\lim_n f(n)= \lim_x f(x) \,.$$

I don't recall if $\left(1+\frac{1}{x}\right)^x$ is monotonic (and I am too lazy to differentiate it), but what we showed above is that it is at least "close to being monotonic". What I mean by this, I showed that $$f(n) h(n) \leq f(x) \leq f(n+1)g(n+1) \forall x\in [n, n+1)$$ where $h(n)$ and $g(n)$ go to $1$. The you basically sqeeze it.

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    $\begingroup$ $(1+\frac{1}{x})^x$ increases monotonicly for positive $x$ $\endgroup$ – Henry Oct 20 '11 at 23:28
  • $\begingroup$ @DJC: Good idea, but since both $\log\left(1+\frac{1}{x}\right)$ and $\frac{1}{x+1}$ are less than $\frac{1}{x}$, I think you need to go a bit farther in the asymptotic expansion: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\right)$ and $\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$. And even then, we need to consider when $O\left(\frac{1}{x^3}\right)$ is small enough. $\endgroup$ – robjohn Oct 21 '11 at 0:16
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    $\begingroup$ Since $e^u\ge1+u$ for all $u\in\mathbb{R}$, we also have $u\ge\log(1+u)$ for $u>-1$. The derivative of the logarithm of $(1+1/x)^x$ is $$ \begin{align} \log\left(1+\frac{1}{x}\right)-\frac{1}{x+1} &=-\log\left(1-\frac{1}{x+1}\right)-\frac{1}{x+1}\\ &=u-\log(1+u)\\ &\ge0 \end{align} $$ where $u=-\frac{1}{x+1}>-1$ for $x>0$. Therefore, $(1+1/x)^x$ is monotonically increasing for $x>0$. $\endgroup$ – robjohn Oct 21 '11 at 0:41
  • $\begingroup$ Your sandwich between $\left(1+\frac{1}{n+1} \right)^n$ and $\left(1+\frac{1}{n}\right)^{n+1}$ is similar to this sandwich. (+1) $\endgroup$ – robjohn Oct 21 '11 at 2:13
  • $\begingroup$ @robjohn: You're absolutely right. Thanks for cleaning that up. $\endgroup$ – JavaMan Oct 21 '11 at 4:03

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