Prove $\lim_{n \to \infty}{\frac{e^{h_n} - 1}{h_n}} = 1$, where $h_n \gt 0$ and $\lim_{n \to \infty}h_n = 0$. $\displaystyle\lim_{n \to \infty}{\frac{e^{h_n} - 1}{h_n}} = 1$ where $h_n \gt 0$ and $\lim_{n \to \infty}{h_n} = 0$.
I would like to see this done using the fact that $\displaystyle\lim_{n \to \infty}{\frac{e^{\frac{1}{n}} - 1}{\frac{1}{n}}} = 1$
I had a couple of ideas but they kind of fell through and didn't work.
I appreciate any and all help.
 A: Let $y_n=e^{h_n}-1$. Then $\frac{e^{h_n}-1}{h_n}=\frac{y_n}{\ln(y_n+1)}=\frac{1}{\frac{1}{y_n}\ln(y_n+1)}=\frac{1}{\ln(y_n+1)^{1/{y_n}}}$. What happens as $n\to \infty$? Are you allowed to assume $\ln$ is continuous and that limits can be brought inside continuous functions?
A: Let $\varphi(x)=\frac{e^x-1}{x}$, then $\forall x>0,\varphi'(x)=\frac{e^x(x-1)+1}{x^2}\geqslant0$ because $e^x(x-1)\geqslant x-1\geqslant -1$. Thus $\varphi$ is non-decreasing. Notice also that $\varphi(x)\geqslant 1$ for all $x>0$ (because $e^x\geqslant x+1$ for all $x\in\mathbb{R}$). Let $\varepsilon>0$ and $m\in\mathbb{N}$ such that $\left|\varphi\left(\frac{1}{m}\right)-1\right|<\varepsilon$, since $\lim\limits_{n\rightarrow +\infty}h_n=0$, there exists $p$ such that $\forall n\geqslant p,0<h_n\leqslant\frac{1}{m}$. Now for all $n\geqslant p$, $0<h_n\leqslant\frac{1}{m}$ and since $\varphi$ is non-decreasing we have $1\leqslant\varphi(h_n)\leqslant\varphi\left(\frac{1}{m}\right)<1+\varepsilon$ which means that $\lim\limits_{n\rightarrow +\infty}\varphi(h_n)=1$.
A: It is not harder than showing that 
$$\frac{e^{1/n}-1}{1/n} \to 1$$
which uses 
$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n-1}\right)^n$
that is 
$\frac{1}{n} < e^{\frac{1}{n}}-1 < \frac{1}{n-1}$
So assuming that you know the above limit, take $h>0$ small and place it between two fractions $\frac{1}{n}\le h \le \frac{1}{n-1}$. Then 
$$\frac{e^{1/n} -1}{1/(n-1)} \le \frac{e^h-1}{h}\le \frac{e^{1/(n-1)}-1}{1/n}$$
Now LHS $=\frac{e^{1/n} -1}{1/n}\cdot (n-1)/n$ and
RHS $=\frac{e^{1/(n-1)}-1}{1/(n-1)}\cdot n/(n-1)$
and you can use the above limit and the squeeze theorem. 
Note this determines the (right) derivative of the function $e^x$ at $x=0$. 
