Chapter 2.3, Problem 29E

If 4x – 9 ≤ f(x)≤ x² – 4x + 7 for x ≥ 0, find $\underset{x\to 4}{\lim }f(x)$

To determine

To evaluate: The limit value of the function $f\left(x\right)$ as x approaches 4.

Answer to Problem 29E

The limit value of the function $f\left(x\right)$ as x approaches 4 is 7.

Explanation of Solution

Given:

The inequality, $4x-9\le f\left(x\right)\le x^2-4x+7$ for $x\ge 0$.

Limit Laws:

Suppose that c is a constant and the limits $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exist, then

Limit law 1: $\underset{x\to a}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)$

Limit law 2: $\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)$

Limit law 3: $\underset{x\to a}{\lim }\left[cf\left(x\right)\right]=c\underset{x\to a}{\lim }f\left(x\right)$

Limit law 7: $\underset{x\to a}{\lim }c=c$

Limit law 8: $\underset{x\to a}{\lim }x=a$

Limit law 9: $\underset{x\to a}{\lim }{x}^n=a^n$ where n is a positive integer.

Theorem used: The Squeeze Theorem

“If $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ when x is near a (except possibly at a) and $\underset{x\to a}{\lim }g\left(x\right)=\underset{x\to a}{\lim }h\left(x\right)=L$ then $\underset{x\to a}{\lim }f\left(x\right)=L$.”

Calculation:

Apply the Squeeze Theorem and obtain a function g smaller than $f\left(x\right)$ and a function h bigger than $f\left(x\right)$ such that both $g\left(x\right)$ and $h\left(x\right)$ approaches 4.

The given inequality becomes, $4x-9\le f\left(x\right)\le x^2-4x+7$.

When the limit x approaches to 4, the inequality becomes,

$\underset{x\to 4}{\lim }4x-9\le \underset{x\to 4}{\lim }f\left(x\right)\le \underset{x\to 4}{\lim }{x}^2-4x+7$

Let $g\left(x\right)=4x-9$ and $h\left(x\right)=x^2-4x+7$.

Obtain the limit value of as x approaches 4 of the function $g\left(x\right)$.

$\begin{array}{c}\underset{x\to 4}{\lim }4x-9=\underset{x\to 4}{\lim }4x-\underset{x\to 4}{\lim }9\left(\text{by limit law 2}\right)\\ =4\underset{x\to 4}{\lim }x-\underset{x\to 4}{\lim }9\left(\text{by limit law 3}\right)\\ =4\left(4\right)-\left(9\right)\left(\text{by limit law 8 and 7}\right)\\ =7\end{array}$

Obtain the limit value of as x approaches 4 of the function $h\left(x\right)$.

$\begin{array}{c}\underset{x\to 4}{\lim }{x}^2-4x+7=\underset{x\to 4}{\lim }\left(x^2-4x\right)+\underset{x\to 4}{\lim }7\left(\text{by limit law 1}\right)\\ =\underset{x\to 4}{\lim }\left(x^2\right)-\underset{x\to 4}{\lim }4x+\underset{x\to 4}{\lim }7\left(\text{by limit law 2}\right)\\ =\underset{x\to 4}{\lim }\left(x^2\right)-4\underset{x\to 4}{\lim }x+\underset{x\to 4}{\lim }7\left(\text{by limit law 3}\right)\end{array}$

                        $\begin{array}{c}={\left(4\right)}^2-4\left(4\right)+\left(7\right)\left(\text{by limit law 9,8 and 7}\right)\\ =16-16+7\\ =7\end{array}$

If $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ when x approaches 4 and $\underset{x\to 4}{\lim }4x-9=\underset{x\to 4}{\lim }{x}^2-4x+7=7$, then by Squeeze Theorem the limit of the function $f\left(x\right)$ is 7.

That is, $\underset{x\to 4}{\lim }f\left(x\right)=7$.

Thus, the limit value of the function $f\left(x\right)$ as x approaches 4 is 7.