How to show that $\lim_{n\rightarrow \infty }(1+\frac{x}{n})^{n}=e^{x}$

Question

I have a problem:

$E(x)=e^x$, $L(x)=\ln (x)$, $E^{-1}(x)=L(x)$.

Show that $\lim_{n\rightarrow \infty }(1+\frac{x}{n})^{n}=e^{x}$

Hint: use $f(t)=\ln (1+xt)$ and look at $f'(0), x\neq 0$.

I found out that $f'(0)=x$, so I can put it in $e^{f'(0)}$, but I don't know how it can help me. Please, can anyone here help me with this problem or give me a link to a good proof. I'm desperate.

Answer 1

Use that $\ln$ is continuous, so show that $$\lim_{n\to\infty} \ln((1+\frac xn)^n) = x$$ Where $$\ln((1+\frac xn)^n) = n\ln(1+\frac xn) = \frac{f(\frac1n)-f(0)}{\frac1n} \to f'(0) = x$$ By the definition of the derivative.

Answer 2

you already have $f'(0) = x$ for $f(x) = \ln(1+xt)$ if we write the definition of $$ x = f'(0) = \lim_{n \to \infty} \frac{f(1/n) - f(0)}{1/n} = \lim_{n \to \infty}n\left(\ln(1 + x/n)\right) = \lim_{n \to \infty}\ln((1+x/n)^n$$

exponentiating the equality gives you $$e^x = \lim_{n \to \infty}\left(1+\frac xn\right)^n. $$