How to rewrite $\lim_{h \to 0}\frac{e^h - 1}{h}=1$ into $\lim_{n \to +\infty}\left(\frac{x}{n}+1\right)^n=e^x$ I am trying to rewrite $\lim_{h \to 0}\frac{e^h - 1}{h}=1$ into $\lim_{n \to +\infty}\left(1 + \frac{x}{n}\right)^n=e^x$.
Developing:
$$\lim_{h \to 0}\frac{e^h - 1}{h}=1$$
$$\lim_{h \to 0}1 + h=\lim_{h \to 0}e^h$$
$$\lim_{h \to 0}\left(1 + h\right)^{\frac{1}{h}}=e$$
Changing the limit variable:
$$\lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^{n}=e$$
Raising to the $x$-power and simplifying by the limits law of powers:
$$\left(\lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^{n}\right)^x=e^x$$
$$\lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^{nx}=e^x$$
I am stuck here, I do not understand how I can substitute this $\frac{1}{n}$ into $\frac{x}{n}$ and remove the $x$-power. I am not at ease with limits and there must be something I am missing.
 A: Substituting $m=nx$ we have $n=\frac{m}{x}$ and for $x>0$ we get 
$$\lim_{m \to +\infty}\left(1 + \frac{x}{m}\right)^{m}$$
$x=0$ is obvious, if $x<0$ put $y=-x$ and we get
$$\lim_{n \to +\infty} \frac{1}{(1 + \frac{1}{n})^{ny}}$$
and then it can be treated as before.
A: Your approach seems like a mix of algebraic manipulation and calculus stuff. This is wrong. Limits can be manipulated but using their own set of laws which are not exactly the same as typical rules one learns in high school algebra.
Your step $$\lim_{h\to 0}1+h=\lim_{h\to 0}e^h\implies \lim_{h\to 0}(1+h)^{1/h}=e$$ is clearly false. The first part of implication is true as both sides equal $1$ but it does not imply the second equation. This is because the first part remains true even if one replaces $e$ by any positive number. Following that implication it appears that $\lim_{h\to 0}(1+h)^{1/h}$ can take any positive value.
Your steps appear like algebraic manipulation $$\frac{e^h-1}{h}=1\implies 1+h=e^h\implies (1+h)^{1/h}=e$$ without the limit operator involved. Here the logical flow is correct but the equations are false individually. The equations in your approach are true but they don't follow logically from one another (at least not via algebraic manipulation). 
Here is an even more interesting scenario :
$$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n=y\implies\lim_{n\to\infty} n(y^{1/n}-1)=x$$ and $$\left(1+\frac{x}{n}\right)^n=y\implies n(y^{1/n}-1)=x$$ Both the implications are true, the second one being just a matter of simple algebra. The first one is an implication with many hidden steps. 

The correct and rigorous approach starts with a proper definition of symbol $a^b$ with $a>0,b\in\mathbb {R} $ and a study of the function $f(x) =a^x$ where $a>0$. One can prove that this function is continuous everywhere and its derivative is $$f'(x) =a^x\lim_{h\to 0}\frac{a^h-1}{h}=a^xg(a)$$ where the limit above exists for all positive $a$ and hence defines a functions $g:\mathbb{R} ^{+} \to \mathbb{R} $.
Next one starts a study of new function $g$ and learns that $g$ is continuous and strictly increasing on its domain. Further it satisfies the following properties $$g(ab) =g(a) +g(b), \lim_{h\to 0}\frac{g(1+h)}{h}=1$$ Here is how one derives the first property mentioned above (at the request of asker via comments) :
\begin{align} 
g(ab) &=\lim_{h\to 0}\frac{(ab)^h-1}{h}\notag\\
&=\lim_{h\to 0}\frac{a^hb^h-b^h+b^h-1}{h}\notag\\
&=\lim_{h\to 0}b^h\cdot\frac{a^h-1}{h}+\frac{b^h-1}{h}\notag\\
&=1\cdot g(a) +g(b) \notag\\
&=g(a) +g(b) \notag
\end{align} 
Using these properties one can easily show that $g$ is differentiable with $g'(x) =1/x$.
Another key property of $g$ is $g(a^b) =bg(a) $ for all real $b$ and this allows us to conclude that range of $g$ is whole of $\mathbb {R} $. By the strict monotone nature of $g$ it follows that there is a unique real number $e>1$ such that $g(e) =1$. And then we get $$g(e^x) =xg(e) =x$$ so that $g$ is the inverse of the specific exponential function $\exp(x)=e^x$.
We can now prove more generally that $$\lim_{h\to 0}(1+xh)^{1/h}=e^x$$ To proceed we start with the limit $$\lim_{h\to 0}\frac{g(1+xh)}{h}=x$$ or $$\lim_{h\to 0}g((1+xh)^{1/h})=x$$ and by continuity of exponential function $\exp$ one gets $$\lim_{h\to 0}\exp(g((1+xh)^{1/h})) =\exp (x) =e^x$$ or $$\lim_{h\to 0}(1+xh)^{1/h}=e^x$$ Putting $h=1/n$ we get the desired limit formula. 
A: Substituting $$t=e^h-1$$ we get $$h=\ln(t+1)$$ and our term is given by $$\frac{t}{\ln(t+1)}=\frac{1}{\ln(1+t)^{1/t}}$$ so our searched limit is given by $1$ if $t$ tends to zero.(If $h$ tends to zero so $t$ tends to zero).
