Question

The figure that follows at the left shows the graphs of $y=\sinh x,y=\cosh x,\text{and}y=\frac{1}{2}{e}^x.$For each of the statements that follows, explain graphically why the statement is true. Then justify the statement algebraically, using the definitions of the hyperbolic functions.

$$\begin{gathered} \text{(a)}\sinh x\le \frac{1}{2}{e}^x\le \cosh x\text{for all}x \\ \text{(b)}\underset{x\to \infty }{\lim }\frac{\sinh x}{(1/2)e^x}=1\text{and}\underset{x\to \infty }{\lim }\frac{\cosh x}{(1/2)e^x}=1 \end{gathered}$$

Short answer

The expressions are proved.

Step-by-step solution

Part (a) Step 1. Given information.

The given expression is$\sinh x\le \frac{1}{2}{e}^x\le \cosh x\text{for all}x$.

Part (a) Step 2. Explanation.

We know,

$\frac{1}{2}{e}^{-x}\ge 0\text{for all}x$

Therefore,

On splitting the terms,

$\sinh x\le \frac{1}{2}{e}^x\le \cosh x\text{for all}x$

Hence proved.

Part (b) Step 1. Explanation.

Calculating the two limits by dividing top and bottom by, it is ,

$\underset{x\to \infty }{\lim }\frac{\sinh x}{(1/2)e^x}=1$

$\underset{x\to \infty }{\lim }\frac{\cosh x}{(1/2)e^x}=1$

Even from the graph,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{\sinh x}{(1/2)e^x}=1 \\ \underset{x\to \infty }{\lim }\frac{\cosh x}{(1/2)e^x}=1 \end{gathered}$$