Chapter 15.1, Problem 30E

To determine

To find:

The value of given limit at thatpoint .

Answer to Problem 30E

  $\underset{x\to 4}{\lim }\frac{2x-8}{x^3-64}=\frac{1}{24}$

Explanation of Solution

Given:

The function $f$ is

  $\underset{x\to 4}{\lim }\frac{2x-8}{x^3-64}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

  $\underset{x\to a}{\lim }f\left(x\right)=L$

Calculation:

The existence of the limit

  $\begin{array}{l}\underset{x\to 4}{\lim }\frac{2x-8}{x^3-64}\\ \text{factor}\\ x^3-64\\ \left(x-4\right)\left(x^2+16+4x\right)\\ \underset{x\to 4}{\lim }\frac{2x-8}{\left(x-4\right)\left(x^2+16+4x\right)}\\ \underset{x\to 4}{\lim }\frac{2}{\left(x^2+16+4x\right)}\end{array}$

Apply by limit

  $\begin{array}{l}=\underset{x\to 4}{\lim }\frac{2}{\left(x^2+16+4x\right)}\\ =\frac{2}{\left(4^2+16+4\left(4\right)\right)}\\ =\frac{2}{48}\\ =\frac{1}{24}\end{array}$

The value of $\underset{x\to 4}{\lim }\frac{2x-8}{x^3-64}=\frac{1}{24}$ .