Limit of $(1+ x/n)^n$ when $n$ tends to infinity

A short proof:

$\left(1+\frac{x}{n}\right)^n = e^{n\log\left(1+\dfrac{x}{n}\right)}$

Since $\log(1+x) = x + O(x^2)$ when $x \to 0$, we have $n\log(1 + \frac{x}{n}) = x + O(\frac{x^2}{n})$ when $n\to +\infty$

$$e^{\ln{(1 + \frac{x}{n})^n} }=e^{n \ln(1+\frac{x}{n})}$$

$$\lim_{n \to +\infty} (1 + \frac{x}{n})^n =\lim_{n \to +\infty} e^{n \ln(1+\frac{x}{n})} \\ =e^{\lim_{n \to +\infty} n \ln(1+\frac{x}{n})} =e^{\lim_{n \to +\infty}\frac{ \ln(1+\frac{x}{n})}{\frac{1}{n}}}$$

Apply L'Hopital's Rule:

$$=e^{\lim_{n \to +\infty}\frac{(\frac{-x}{n^2})\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}}} =e^{\lim_{n \to +\infty}\frac{x}{1+\frac{x}{n}}} =e^x$$

Therefore, $$(1+\frac{x}{n})^n \to e^x$$

You can use the binomial series expansion. For example:

$$\left(1+\frac{x}{n}\right)^n =1+ \frac{n}{1!}\left(\frac{x}{n}\right)^1+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots $$

$$\left(1+\frac{x}{n}\right)^n =1+ \frac{n}{n}x+\frac{n(n-1)}{n^2}\frac{x^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{x^3}{3!} + \cdots$$

As $n \to \infty$ the coefficients in $n$ all tend to $1$. Hence:

$$\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots $$ You'll recognise this last power series as the Taylor series for $\mathrm{e}^x$.