Chapter 2.2, Problem 20E

To determine

To guess: The value of $\underset{h\to 0}{\lim }\frac{{(2+h)}^5-32}{h}$.

Answer to Problem 20E

The value of limit is guessed to 80.

Explanation of Solution

Given:

The values of h are $h=\pm 0.5,\pm 0.1,\pm 0.01,\pm 0.001,\text{ and }\pm 0.0001$.

Calculation:

Since h approaches 0, the values $-0.5,-0.1,-0.01,-0.001,\text{ and }-0.0001$ are to the left of 0 and the values $0.5,0.1,0.01,0.001,\text{ and }0.0001$ are to the right of 0.

Evaluate the function (correct to 6 decimal places) when $h=-0.5,-0.1,-0.01,-0.001,\text{ and }-0.0001$ as shown in the below table.

$h$	${\left(2+h\right)}^5$	${\left(2+h\right)}^5-32$	$f\left(h\right)=\frac{{(2+h)}^5-32}{h}$
−0.5	7.59375	−24.4063	48.812500
−0.1	24.76099	−7.23901	72.390100
−0.01	31.20796	−0.79204	79.203990
−0.001	31.92008	−0.07992	79.920040
−0.0001	31.992	−0.008	79.992000

Here, the value of $f\left(h\right)$ approaches 80 as h closer to 0 from the left side.

That is, $\underset{h\to 0^{-}}{\lim }\frac{{(2+h)}^5-32}{h}=80$.

Evaluate the function (correct to 6 decimal places) when $h=0.5,0.1,0.01,0.001,\text{ and }0.0001$ as shown in the below table.

$h$	${\left(2+h\right)}^5$	${\left(2+h\right)}^5-32$	$f\left(h\right)=\frac{{(2+h)}^5-32}{h}$
0.5	97.65625	65.65625	131.312500
0.1	40.84101	8.84101	88.410100
0.01	32.80804	0.80804	80.804010
0.001	32.08008	0.08008	80.080040
0.0001	32.008	0.008001	80.008000

Here, the value of $f\left(h\right)$ approaches 80 as h closer to 0 from the right side.

That is, $\underset{h\to 0^{+}}{\lim }\frac{{(2+h)}^5-32}{h}=80$.

Since the right hand limit and the left hand limits are the same, the value of $\underset{h\to 0}{\lim }\frac{{(2+h)}^5-32}{h}$ exists.

Therefore, the limit of the function is guessed to 80.