Chapter 12, Problem 5CT

To determine

To graph: The given function, to approximate its limit, to create its table of values and to verify the approximation numerically.

Answer to Problem 5CT

  $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}=2.$

Explanation of Solution

Given Information: Given limit is $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}$ .

Graph:

The graph of given function $\frac{e^{2x}-1}{x}$ is:

  

Let us choose two points on the graph, say, $(-0.0004,1.9992)and(0.0004,2.0008)$ . The $x-coordinates$ of these two points are equidistant from $0$ , so approximating the limit to the average of the $y-coordinate$ , i.e.,

  $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}$

  $=\frac{1.9992+2.0008}{2}=2.$

Verification:

  $\begin{array}{llllllllll}x\hfill & -0.0004\hfill & -0.0003\hfill & -0.0002\hfill & -0.0001\hfill & 0\hfill & 0.0001\hfill & 0.0002\hfill & 0.0003\hfill & 0.0004\hfill \\ f(x)\hfill & 1.9992\hfill & 1.9994\hfill & 1.9996\hfill & 1.9998\hfill & undefined\hfill & 2.0002\hfill & 2.0004\hfill & 2.0006\hfill & 2.0008\hfill \end{array}$

Because $0$ is halfway between, $-0.0001and0.0001$ , so using the average of the values of $f$ at these two $x-values$ to estimate the limit,

  $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}=\frac{1.9998+2.0002}{2}=2.$

Hence verified.

Thus, $\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}=2.$