Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$ 
Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy. 
But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

 A: Start with $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$ This means that $$e^x - 1= x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$
and that $$\frac{e^x - 1}x= 1+\frac {x}{2!}+\frac {x^2}{3!}+\ldots$$
I believe you can take it from here.
A: Another way to do it is to define $f: \mathbb{R} \to \mathbb{R}$ by setting $f(x)=e^x$, then note that $f(0) = 1$, in that case the derivative of $f$ at the point $0$ can be given by the limit:
$$f'(0)=\lim_{x\to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{e^x - 1}{x}$$
In that case your limit is just $f'(0)$ which is $1$ since $f'(0) = e^0$. Of course this only works when you know that $f'(x) = e^x$. This helps you somehow?
A: Use L'Hopital's rule: 
$$\frac{d}{dx}(e^x -1 ) = e^x$$ 
And the denominator is $1$. Filling with the limit, gives $e^0/1 = 1$.
A: Take $f:\mathbb R-\{0\}\to\mathbb R:x\mapsto e^x-1\\g:(-1,\infty)-\{0\}\to\mathbb R:x\mapsto\dfrac{x}{\log(x+1)}.$ 
Note $\forall~x\in\mathbb R,~e^x>0\text{ i.e. } e^x-1>-1\text{ and }e^x=1\iff x=0.\text{ So image}(f)\subset\text{domain}(g).$ Now $gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto \dfrac{e^x-1}{x}.$
The given limit is $\displaystyle\lim_{x\to0}(gf)(x).$ $e^x-1$ being continuous at $0,\displaystyle\lim_{x\to0}f(x)=0,$ a limit point of the domain$(g)$ and $\displaystyle\lim_{x\to0}\frac{\log(1+x)}{x}=1(\neq 0)\implies\displaystyle\lim_{x\to0}\frac{x}{\log(1+x)}=1.$ Hence the given limit is $\displaystyle\lim_{x\to0}g(x)=1.$
A: I would do this by using L'Hopitals rule.  Because $\lim_{x \to 0} \frac{e^{x}-1}{x}= \frac{0}{0}$, we can take the the derivative of the numerator and denominator individually, then plug in 0 again to the new equation.  After taking the derivative of the numerator and denominator we will have $\lim_{x \to  0}\frac{e^x}{1}$, and from here we see why the answer is $1$.
