Chapter 1, Problem 56EP

(a)

To determine

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

Answer to Problem 56EP

The value of $\underset{x\to 2^{-}}{\lim }f(x)=4-2a^2$

Explanation of Solution

Given information: The function is $f(x)=\left\{\begin{array}{c}{x}^2-a^{2}x\text{ if }x<2\\ 4-2x^2\text{if }x\ge 2\end{array}\right.$

Calculation:

The left side limit of $f(x)$ ,use the definition of $f$ for $x<2$ :

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

Therefore, the required limit is $\underset{x\to 2^{-}}{\lim }f(x)=4-2a^2$ .

(b)

To determine

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

Answer to Problem 56EP

The value of $\underset{x\to 2^{+}}{\lim }f(x)=-4$

Explanation of Solution

Given information: The function is $f(x)=\left\{\begin{array}{c}{x}^2-a^{2}x\text{ if }x<2\\ 4-2x^2\text{if }x\ge 2\end{array}\right.$

Calculation:

The portion of the function $f(x)=4-2x^2$ because in order to determine the right hand limit, we must look at the integers $x\ge 2$ and calculate,

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

Therefore, the required value of the limit is $\underset{x\to 2^{+}}{\lim }f(x)=-4$ .

(c)

To determine

To find: The all values of $a$ that make $f$ continuous at 2.

Answer to Problem 56EP

The value of $a=\pm 2$ .

Explanation of Solution

Given information: The function is $f(x)=\left\{\begin{array}{c}{x}^2-a^{2}x\text{ if }x<2\\ 4-2x^2\text{if }x\ge 2\end{array}\right.$

Calculation:

From part 56 (a) and part 56 (b):

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

For $f$ to be continuous, these two limits must be equal,

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

Therefore, the required values of $a$ that make $f$ continuous at 2 is $a=\pm 2$ .