Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

\(\begin{array}{l}\mathop {lim}\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^5} - 32}}{h}\\h = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001\end{array}\)

Short answer

The value of the limit is 80.

Step-by-step solution

Introduction

The function does not exist on the given limit but it can sustain on the nearby values of the limit. To find the value of the limit we have to find the nearby limit values.

Given information

The given function and the values are,

\(\begin{array}{l}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^5} - 32}}{h}\\h = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001\end{array}\)

Explanation

The following chart is for\(\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^5} - 32}}{h}\)with the values\(h = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001\).

x	f(x)	x	f(x)
0.5	131.312500	\( - 0.5\)	48.812500
0.1	88.410100	\( - 0.1\)	72.390100
0.01	80.804010	\( - 0.01\)	79.203990
0.001	80.080040	\( - 0.001\)	79.920039
0.0001	80.008000	\( - 0.0001\)	79.992000

Hence the value of the limit is 80.