I want to show that
$$\lim_{h \to 0} \frac{\text{e}^h -1}{h}=1$$
by using the Squeeze theorem. Is it possible to prove this with the Squeeze theorem?
Maybe the two inequalities
$$ \forall \, h \in (-1,1) \colon 1+h \leq \text{e}^h \leq \frac{1}{1-h} $$ $$ \forall \, h > 0 \colon 1 \leq \frac{\text{e}^h-1}{h} \leq \frac{1}{1-h}$$
are helpful. The second inequality looks already good, but the problem is that it holds only for $h > 0$.
Ok, here goes:
For all $x \in \mathbb{R}$ let $n \in \mathbb{N}$ such as $n>-x$. Hence $\dfrac{x}{n}+1>0$, so $\displaystyle \left ( 1+\frac{x}{n} \right )^n\geq 1+n \cdot \frac{x}{n}=1+x$. (from Bernoulli's inequality)
Therefore: $$e^x=\lim \left ( 1+\frac{x}{n} \right )^n \geq \lim (1+x) =1+x \implies e^x \geq 1+x$$
Subbing in the previous equation whereas $x$ the $-x$ we get: $e^{-x}\geq 1-x$ and forall $x \in (-1, 1)$ we have $\displaystyle e^x\leq \frac{1}{1-x}$.
Thus we get: $$1+x\leq e^x\leq \frac{1}{1-x} \implies x\leq e^x-1\leq \frac{x}{x-1}$$ (which is what you've got).Now divide by $x$
If $0<x<1$ then the inequality remains as it is, otherwise if $-1<x<0$ the inequality is reversed. In both cases using the squeeze theorem you get what you want.