Chapter 2.1, Problem 24E

To determine

To check: Whether the direct substitution can be used for the limit $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ or not.

Answer to Problem 24E

The direct substitution cannot be used for the limit $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ as the function $\frac{{\left(4+x\right)}^2-16}{x}$ is not defined at $x=0$ . The value of $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ is equal to $8$ .

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ .

Calculation:

Substitute $0$ for $x$ in the function.

  $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}=\frac{{\left(4+0\right)}^2-16}{0}$

This is not defined.

Simplify the given limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}=\underset{x\to 0}{\lim }\frac{16+x^2+8x-16}{x}\\ =\underset{x\to 0}{\lim }\frac{x\left(x+8\right)}{x}\\ =\underset{x\to 0}{\lim }\left(x+8\right)\\ =8\end{array}$

So, the value of $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ is equal to $8$ .

Therefore, the value of $\underset{x\to 0}{\lim }\frac{{\left(4+x\right)}^2-16}{x}$ exist and is equal to $8$ . The direct substitution cannot be used as the function is not defined at $x=0$ .