proof $|\sinh(x)|\leq|3x|$ for $ -\frac{1}{2}I am supposed to prove that $|\sinh(x)|\leq3|x|$ for $|x|<\frac{1}{2}$. 
I know I am supposed to use $|e^x-1|\leq3|x|$ for $|x|<\frac{1}{2}$. 
I am completely stuck, and I don't know how to approach this, so any help is greatly appreciated!

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities 

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le\frac{1}{1-x}} \tag 1$$

for $x<1$.

Let $f(x) = 6x-e^x+e^{-x}$.  Then, applying $(1)$ to $f(x)$, we find
$$\begin{align}
6x-e^x+e^{-x}&\ge 6x-\frac{1}{1-x}+(1-x)\\\\
&=\frac{x}{1-x}(4-5x)\\\\
&\ge 0
\end{align}$$
for $0\le x\le 4/5$.  

Hence, 
$$\bbox[5px,border:2px solid #C0A000]{|\sinh(x)|\le |3x|}$$
for $|x|\le 4/5$ as was to be shown!

