Chapter 2, Problem 54RE

(a)

To determine

To find: The value of ${\lim }_{x\to 2^{-}}f\left(x\right)$

Answer to Problem 54RE

The value of ${\lim }_{x\to 2^{-}}f\left(x\right)$ is $4-2a^2$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}{x}^2-a^{2x},x<2\\ 4-2x^2,x\ge 2\end{array}$ .

Calculation:

The left-hand limit of the function at $x=2$ is:

  $\begin{array}{c}{\lim }_{x\to 2^{-}}f\left(x\right)={\lim }_{x\to 2^{-}}\left(x^2-a^{2}x\right)\\ =4-2a^2\end{array}$

Therefore, thevalue of ${\lim }_{x\to 2^{-}}f\left(x\right)$ is $4-2a^2$ .

(b)

To determine

To find:The value of ${\lim }_{x\to 2^{+}}f\left(x\right)$

Answer to Problem 54RE

The value of ${\lim }_{x\to 2^{+}}f\left(x\right)$ is $-4$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}{x}^2-a^{2x},x<2\\ 4-2x^2,x\ge 2\end{array}$ .

Calculation:

The right-hand limit of the function at $x=2$ is:

  $\begin{array}{c}{\lim }_{x\to 2^{+}}f\left(x\right)={\lim }_{x\to 2^{+}}\left(4-2x^2\right)\\ =\left(4-8\right)\\ =-4\end{array}$

Therefore, the value of ${\lim }_{x\to 2^{+}}f\left(x\right)$ is $-4$ .

(c)

To determine

To find: The values of a that make the given function continuous at 2.

Answer to Problem 54RE

The value of a that make the given function continuous are 2 and $-2$

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}{x}^2-a^{2x},x<2\\ 4-2x^2,x\ge 2\end{array}$ .

Calculation:

Calculate the function.

  $\begin{array}{c}{\lim }_{x\to 2^{+}}f\left(x\right)={\lim }_{x\to 2^{-}}f\left(x\right)\\ =f\left(2\right)\end{array}$

To make the function continuous, the left-hand limit should be equal to the right-hand limit. So,

  $\begin{array}{c}4-2a^2=-4\\ 2a^2=8\\ a^2=4\\ a=2,-2\end{array}$

Therefore, the value of a that make the given function continuous are 2 and $-2$ .