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?
 A: 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
$$\lim_{n\to \infty}\left(1+\frac{1}{k}\right)^{kx}= \left(\lim_{k\to \infty}\left(1+\frac{1}{k}\right)^{k}\right)^x =e^x$$ 
A: 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 $f(x) = e^x$.
A: Let's start with the "where did I go wrong" part of the question.  Where you wrote
$$=n\left[\int_1^{n}\frac{dt}{t}+\int_{n}^{x}\frac{dt}{t}-\int_1^{n}\frac{dt}{t}\right]$$
you should have written
$$=n\left[\int_1^{n}\frac{dt}{t}+\int_{n}^{n+x}\frac{dt}{t}-\int_1^{n}\frac{dt}{t}\right]$$
Note, the correct upper limit in the middle integration is $n+x$, not just $x$.  Otherwise you were on the right track.  The corrected integral leaves us with
$$n\ln\left(1+\frac{x}{n}\right)=n\int_n^{n+x}{dt\over t}$$
Now as soon at $n\gt x$, we have
$${1\over n+|x|}\le{1\over t}\le{1\over n-|x|}$$
on the interval $t\in[n-|x|,n+|x|]$, which certainly includes the interval between $n$ and $n+x$.  If you are careful with the minus signs, you can conclude that
$${nx\over n+x}\le n\int_n^{n+x}{dt\over t}\le{nx\over n-x}$$
and it now follows easily that
$$\lim_{n\to\infty}n\ln\left(1+\frac{x}{n}\right)=x$$
A: If you're allowed to use Taylor (power) expansions this is pretty simple:
$$n\log\left(1+\frac xn\right)=n\sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k\,n^k}=n\left(\frac xn+\mathcal O\left(\frac1{n^2}\right)\right)=$$
$$=x+\mathcal O\left(\frac1n\right)\xrightarrow[n\to\infty]{}x$$
A: If you already know the exponential function, then you probably also know the inequality $e^x\ge 1+x$. Then also
$$e^x=\frac1{e^{-x}}\le\frac1{1-x}=(1+x)\frac1{1-x^2}$$
Now $e^x=(e^{\frac xn})^n$, so for $n>|x|$
$$
\left(1+\tfrac xn\right)^n
\le e^x
\le \left(1+\tfrac{x}{n}\right)^n\frac1{\left(1-\tfrac{x^2}{n^2}\right)^n}\le\left(1+\tfrac{x}{n}\right)^n\frac1{1-\tfrac{x^2}{n}}
$$
The last inequality is by the Bernoulli inequality.
A: First, let's prove that :
$$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$
For that, we need to know :
$$(2)\space\space\lim_{x \to 0} \frac{ln(1 + x)}{x}$$
Let's say : $\forall x > -1,\space f(x) = ln(1 + x)$
$f(0) = ln(1) = 0$, so :
$$\frac{ln(1+x)}{x} = \frac{f(x) - f(0)}{x} = \frac{f(0+x) - f(0)}{(0 + x) - 0}$$
By using the definition of the derivative,
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{x \to 0} \frac{f(0+x) - f(0)}{(0 + x) - 0} = f'(0)$$
$$\forall x > -1,\space f'(x) = \frac{1}{1+x}\space And\space f'(0) = 1$$
Finally,
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$

Now (2) is computed, let's prove (1) :
Let's say : $\forall x \neq 0, \space y=\frac{1}{x}$
So :
$$x = \frac{1}{y}$$
$x \to 0 \implies y \to \infty$
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{y \to \infty} \frac{ln(1 + \frac{1}{y})}{\frac{1}{y}} = \lim_{y \to \infty} y \space ln\bigg(1 + \frac{1}{y}\bigg)=\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y$$
$$\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = \lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$
$$(3)\space\space\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = 1$$
Let's say exp the inverse function of ln :
$$exp(1) = \lim_{y \to \infty} \bigg(1+\frac{1}{y}\bigg)^y$$
Let's say $e = exp(1)$, and replace y by n, we finally prove :
$$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$

(1) could also be written :
$$(1)\space\space e = \lim_{f(n) \to \infty} \bigg (1+\frac{1}{f(n)} \bigg)^{f(n)}$$
Let's say : $\forall x \neq 0$, $f(n) = \frac{n}{x}$
Let's separate the proof in 3 cases : (a) $x>0$, (b) $x<0$ and (c) $x=0$
Case (a) : $x>0$ so $f(n) \to \infty \implies n \to \infty$
$$e = \lim_{n \to \infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$
By raising at power x :
$$\forall x>0, \space e^x = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
Case (b) : $x<0$ so $f(n) \to \infty \implies n \to -\infty$
$$e = \lim_{n \to -\infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to -\infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$
Let's say $m = -n$
$n \to -\infty \implies m \to \infty$
$$e = \lim_{m \to \infty} \bigg (1+-\frac{1}{\frac{m}{x}} \bigg)^{-\frac{m}{x}} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{\frac{m}{-x}}$$
By raising at power -x :
$$e^{-x} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{m}$$
By replacing -x by x and m by n :
$$\forall x < 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
(1) and (2) implies :
$$\forall x \neq 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
Case (c) : $e \neq 0 \implies e^0 = 1$
Let's see if the formula valid for $x \neq 0$ is also valid for $x=0$ :
$$\space e^{0} = \lim_{n \to \infty} \bigg (1+\frac{0}{n} \bigg)^{n} = \lim_{n \to \infty} 1^n = 1$$
We just proved:
$$\forall x \in \mathbb{R}, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
A: Here's a way to do it with the integral:
\begin{align*}
n\log{\left(1+{x\over n}\right)} & = n\int_1^{1+{x/ n}}{dt\over t}
\end{align*}
and by the simplest conceivable estimate
\begin{align*}
{x\over 1+x/n} = n{x/n\over1+{x/ n}}\leq n\int_1^{1+{x/ n}}{dt\over t} \leq n{x\over n} = x.
\end{align*}
(The inequalities hold when $x$ is negative too, provided the expressions are defined.) Now make $n\to\infty$ and apply the squeeze theorem.
A: $f(x) = e^x$ is the only solution to the differential equation $\dfrac{dy}{dx} = y$ with $f(0)=1$.
To approximate $f(a)$, we can use Euler's method on the interval $[0,a]$ with $n$ subintervals.
$f(0) = 1, f'(0)=1 \implies f(\frac{a}{n}) \approx 1+\frac{a}{n}$
$f(\frac{a}{n}) \approx 1+\frac{a}{n}, f'(\frac{a}{n}) \approx 1+\frac{a}{n} \implies f(\frac{2a}{n}) \approx 1+\frac{a}{n} + \frac{a}{n}(1+\frac{a}{n}) =  (1+\frac{a}{n})^2$
.
.
.
$f(a) \approx (1+\frac{a}{n})^n$
Since Euler's method actually converges in the limit, we have 
$$e^a = \lim_{n \to \infty} (1+\frac{a}{n})^n$$
A: Let's say you define $e^x$ by the derivative $\frac{d}{dx} e^x=e^x$. Then, we try the same to what you have here. $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^{n-1}$. And, as n approaches infinity, n and n-1 essentially boils down to the same thing. Then, $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^n$.
A: Since $X=e^{\log X}$, we can write 
$$\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n=e^{\log\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n}
=e^{\lim_{n \to \infty} n\log \bigg(1+\frac{x}{n}\bigg )}$$
Notice that the exponent on $e$ is an indeterminate form of type $\infty\cdot 0$, so rewrite the limit in the exponent as $\frac{\log \big(1+\frac{x}{n}\big)}{\frac{1}{n}}$ and use L'Hopital's Rule to show the the limit in the exponent of $e$ is indeed $x$.
A: As shown by @math_man, the limit is indeed the exponential of a numerical constant, with
$$\lim \limits_{n\to+\infty }{\left(1+\frac{x}{n}\right)^n}=\lim \limits_{nx\to+\infty }{\left(1+\frac{x}{nx}\right)^{nx}}=\left(\lim \limits_{n\to+\infty }{\left(1+\frac1n\right)^n}\right)^x=e^x.$$
We can compute the constant by the Binomial theorem,
$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}\cdots+\frac1{n^n},$$
which becomes
$$e=1+1+\frac12+\frac1{3!}+\frac1{4!}\cdots$$at infinity.
