Prove $\sinh(x)\leq 3|x|$ for $|x| < 1/2$ I need to prove $|\sinh(x)|\leq 3|x|$ for $|x| < 1/2$
My current progress states that
$$x \leq \frac{ (1+x)-(1-x) }{2} \leq \sinh(x)\leq \frac{ \frac{1}{1-x}-\frac{1}{1+x} }{2}$$
whereas
$$|\sinh(x)|\leq \max\left(|x|,\left|\frac{x}{1+x^2}\right|\right)\leq|x|+\left|\frac{x}{1+x^2}\right|$$
From my book I also get that 
$$|\exp(x)-1|\leq 3|x|\text{ so }|\sinh(x)| \leq |\exp(x)-1|\leq 3|x|$$
But now i'm stuck. Any hints?

It seems you can use the series definition. Then
$$
\left|\sinh(x)\right|=|x|\left(1+\frac{|x|^2}{3!}+\frac{|x|^4}{5!}+…\right)
\le|x|·\left(1+\frac{|x|^2}{6}+\frac{|x|^4}{6^2}+…\right)
\\=|x|·\frac{6}{6-x^2}
$$
for $|x|<\sqrt{6}$ and for $|x|\le\frac12$ this can be reduced to
$$
|\sinh(x)|\le|x|·\frac{24}{23}.
$$
Actually you can extend the range to $|x|<2$ where you get
$$
|\sinh(x)|\le|x|·\frac{6}{6-2^2}=3·|x|.
$$
