Standard limit proof Everyone knows the standard result $$\boxed{\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e}$$ My friend gave a proof for this result using binomial expansion of $$\left(1+\dfrac{1}{x}\right)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\dots$$ which is valid only when $|1/x|<1$ and $n\in\mathbb{R}$. Here is how he proceeded-
$$\begin{align*}\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=\lim_{x\to\infty}\left(1+x.\dfrac1x+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)=\lim_{x\to\infty}\left(1+1+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)=1+1+\lim_{x\to\infty}\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\lim_{x\to\infty}\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots=1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}\dots=\boxed{e}\end{align*}$$ The method of binomial expansion is clearly invalid as the exponent $x$ is variable not a constant. Yet the method ends up getting correct answer as $e$. Am I missing what is incorrect here? I am interested to know why this wrong approach leads to correct answer.
 A: Instead of taking a limit as $x \rightarrow \infty$, one can take a limit along a specific sequence. For instance, the following limit does not exist:
$$\lim_{x\rightarrow \infty}\sin x.$$
Sine is periodic and does not converge to a specific value for large $x$. However, the following limit does exist:
$$\lim_{x=0,\pi,2\pi,\ldots} \sin x = 0.$$
Here we are taking a limit along a countable sequence, skipping over the peaks and troughs of $\sin x$ so that the sine of of every value in the sequence is 0.
Your friend is taking the limit of $(1+\frac{1}{x})^x$ along a particular sequence which grows to infinity, specifically, $x=1,2,3,\ldots$. This is a perfectly valid thing to do, and his math seems correct. Of course, in his case the limit exists not just along this sequence, but as $x \rightarrow \infty$ in general. When the general limit exists, it must equal the limit along every single sequence which grows to infinity, which is what you are seeing here.
A: This is correct for each $\vert x \vert > 1$:
$$ \left( 1+\frac{1}{x} \right) ^ x =
1+1+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots$$
It is simply Newton's Generalised Binomial Theorem with $x$ replaced by $1$, $y$ replaced by $\frac{1}{x}$ and $r$ replaced by $x.$
However, It is not obvious that the following limit even converges (although it probably does for some complicated reason):
$$\lim_{x\to\infty}\left(1+x\cdot\dfrac1x+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)$$
