How to prove $e^x=\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n$ Define
$$e=\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n,$$
how to prove that 
$$e^x=\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$$
for all $x$.
Thank you.
Use the definition

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

Then

$$
e^x = \left( \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n \right)^x
= \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{nx}
= \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{nx} \right)^{nx}
= \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{n} \right)^{n}
$$
By applying $\log$ and  using L'Hospital's rule $$\lim_{n \to \infty}\frac{\log\left(1+\frac{x}{n}\right)}{\frac{1}{n}}=\lim_{n\to \infty}\frac{\frac{1}{1+\frac{x}{n}}\left(\frac{-x}{n^2}\right)}{\frac{-1}{n^2}}=x$$