Proving $\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n=\text{e}^x$.

I knew that $e^x=\lim \limits_{n\to+\infty }{\left(1+\frac{x}{n}\right)^n}$. But I've never seen its proof. So I tried to prove it using $\exp(\ln x)=\ln(\exp(x))=x$. Here is what I've tried so far :

$$ \left(1+\frac{x}{n}\right) ^n=e^{n\ln(1+\frac{x}{n})}$$ $$\text{I'll now study just } {n\ln\left(1+\frac{x}{n}\right)}.$$$$ \text{If this function has the line }y=x \text{ as oblique asymptote, then the equality is proven.}$$

$$ n\ln\left(1+\frac{x}{n}\right) = n\ln\left(\frac{n+x}{n}\right)$$

$$=n[\ln(n+x)-ln(n)]$$ $$=n\left[\int_1^{n}\frac{dt}{t}+\int_{n}^{x}\frac{dt}{t}-\int_1^{n}\frac{dt}{t}\right]$$ $$=n[\ln(x)-\ln(n)]$$

But I just don't know how to show that this expression has an oblique asymptote $y=x$. I've thought that if there is an oblique asymptote as $n$ goes to infinity, than for a huge $n$, we have :

$$\ln\left(1+\frac{x}{n}\right)\approx \frac{x}{n}\approx0$$ Which looks correct but we could have any other function $f(x)$, $\ln\left(1+\frac{x}{n}\right)\approx\frac{f(x)}{n}\approx 0$. Which doesn't prove the oblique asymptote because $x$ is constant.

So how can prove $e^x=\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n$? And where did I go wrong?

Solution 1:

Start with the functions $$ f_n(x) = \left(1 + \frac{x}{n}\right)^n. $$ Then $$ f'_n(x) = \left(1 + \frac{x}{n}\right)^{n-1} = \left(1 + \frac{x}{n}\right)^{-1}f_n(x) $$ If we take the limit and call $f(x) = \lim_{n\rightarrow\infty}f_n(x)$, then $$ f'(x) = f(x) $$ and $f(0) = 1$. This first-order ODE has the unique solution $f(x) = e^x$.

Solution 2:

I don't know if it helps you, it is just a suggestion, if you know the fundamental limite: $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ Then you have for $$\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n$$ replacing $k=\frac{n}{x}$ we get $$\lim_{n\to \infty}\left(1+\frac{1}{k}\right)^{kx}= \left(\lim_{k\to \infty}\left(1+\frac{1}{k}\right)^{k}\right)^x =e^x$$