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

I knew that $e^x=\lim \limits_{n\to+\infty }{(1+\frac{x}{n})^n}$. 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 :

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

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

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

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

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

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

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What is your definition of $e^x$? –  Ink Jun 14 at 6:39
What is your definition of $\exp (x)$? –  Git Gud Jun 14 at 6:40
@moray95: Your integral should be $\int_1^{nx}\frac{dt}{t}+ \int_{nx}^{nx+1} \frac{dt}{t}- \int_1^n \frac{dt}{t}$. –  Seirios Jun 14 at 6:41
I'm using the definition $\exp(\ln(x))=\ln(\exp(x))=x$ –  moray95 Jun 14 at 6:50
Are you defining $\exp$ as the inverse of $\ln$, is that it? –  Git Gud Jun 14 at 6:53

If you're allowed to use Taylor (power) expansions this is pretty simple:

$$n\log\left(1+\frac xn\right)=n\sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k\,n^k}=n\left(\frac xn+\mathcal O\left(\frac1{n^2}\right)\right)=$$

$$=x+\mathcal O\left(\frac1n\right)\xrightarrow[n\to\infty]{}x$$

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Great response but I'd like to prove it without the power series... –  moray95 Jun 14 at 15:51
Ok...did you read my comment under your question? Because you haven't yet corrected your work there... –  DonAntonio Jun 14 at 15:53
Edited my question with your remark but still, I still have the same thing at the end... –  moray95 Jun 14 at 17:07
I don't know if it helps you, it is just a suggestion, if you know the fundamental limite: $$\lim_{n\to \infty}(1+\frac{1}{n})^n=e$$ Then you have for $$\lim_{n\to \infty}(1+\frac{x}{n})^n$$ replacing $k=\frac{n}{x}$ we get $$\lim_{n\to \infty}(1+\frac{1}{k})^{kx}= (\lim_{k\to \infty}(1+\frac{1}{k})^{k})^x =e^x$$
I've thought about something like that but then I'll need to prove first $\lim_{n\to \infty}(1+\frac{1}{n})^n=e$. –  moray95 Jun 15 at 10:23