Other expression of $\exp(x)$

Question

Sorry, I don't know how to use latex.

I heard that $e^x$ equals to $\lim\limits_{n \to \infty} (1+\frac{x}{n})^n$. I tried to prove it, but I couldn't. Why it works and how to make it?

Answer 1

$$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ - The interest rates formula - and so: $$e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}=\lim_{n,m\to\infty}\left(1+\frac{1}{n}\right)^m=\lim_{m\to\infty}\left(1+\frac{1}{m/x}\right)^m=\lim_{m\to\infty}\left(1+\frac{x}{m}\right)^m$$

Where I have set $m=nx$.

A very similar idea is the proof that: $$e^x=\lim_{n\to0}\left(1+xn\right)^{\frac{1}{n}}$$

Try this one for yourself!

Both of these only work for real $x$.

Answer 2

Using the properties of $\ln$ and $\exp$: $$(1+\frac{x}{n})^n=\exp(n \ln(1+\frac{x}{n}))$$ Since $\frac{x}{n} \underset{n \rightarrow + \infty}{\rightarrow} 0$ and $\ln(1+t) \underset{0}{=} t + o(t)$, we have: $$\underset{n \rightarrow + \infty}{\lim} n \ln(1+\frac{x}{n})=n \frac{x}{n}=x$$ And since $\exp$ is continuous, we get $$\underset{n \rightarrow + \infty}{\lim} (1+\frac{x}{n})^n=\exp(x)$$

Answer 3

Show the limit of the log is $x$, which is easy with asymptotic equivalence: $$\ln\Bigl(1+\frac xn\Bigr)^n=n\ln\Bigl(1+\frac xn\Bigr)\sim_\infty n\,\frac xn=x.$$