Algebraic manipulation with $e^x$ I have these inequalities;
$ 1 + x \le e^x \le \frac{1}{1-x} $  for  $|x| <1$
This is what i require;
$ 1 - \frac{|x|}{1-x}  \le e^x \le 1 + \frac{|x|}{1-x}  $  for  $0 < x <1$
This is my attempt;
$ 1 + x \le e^x \le \frac{1}{1-x} $  for  $|x| <1$
$\implies x \le e^x - 1 \le \frac{1}{1-x} -1 = \frac{x}{1-x} $  for  $|x| <1$
$\implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} $  for  $0 < x <1$
I'm not sure how to continue with my argument.

Note that for $x\in(0,1)$
$$
e^x\leq\frac{1}{1-x}=\frac{1-x+x}{1-x}=1+\frac{x}{1-x}=1+\frac{|x|}{1-x}\cdot
$$
This takes care of the the inequality on the right. For the one on the left, (still assuming $x\in(0,1)$)
$$
e^x\geq 1+x>1>1-\underbrace{\frac{|x|}{1-x}}_{+}\cdot
$$
