Chapter 11.1, Problem 53E

a.

To determine

To evaluate $\underset{x\to 2}{\lim }f(x)$ .

Answer to Problem 53E

16

Explanation of Solution

Given:

  $f(x)=x^4$

  $\begin{array}{l}\underset{x\to 2}{\lim }f(x)\\ =\underset{x\to 2}{\lim }{x}^4\\ =2^4\\ =16\end{array}$

b.

To determine

To evaluate. $\underset{x\to 2}{\lim }g(x)$

Answer to Problem 53E

  $\frac{3}{8}$

Explanation of Solution

Given:

  $g(x)=\frac{\sqrt{x^2+5}}{2x^2}$

  $\begin{array}{l}\underset{x\to 2}{\lim }g(x)\\ =\underset{x\to 2}{\lim }\frac{\sqrt{x^2+5}}{2x^2}\\ =\frac{\sqrt{2^2+5}}{2(2^2)}\\ =\frac{\sqrt{9}}{8}\\ =\frac{3}{8}\end{array}$

c.

To determine

To evaluate $\underset{x\to 2}{\lim }f(x)g(x)$

Answer to Problem 53E

6

Explanation of Solution

Given :

  $\begin{array}{l}f(x)=x^4\\ g(x)=\frac{\sqrt{x^2+5}}{2x^2}\end{array}$

  $\begin{array}{l}\underset{x\to 2}{\lim }f(x)g(x)\\ =\underset{x\to 2}{\lim }f(x)\underset{x\to 2}{\lim }g(x)\\ =x^4\frac{\sqrt{x^2+5}}{2x^2}\\ $ ={(2)}^4\frac{\sqrt{2^2+5}}{2{(2)}^2}$=16(\frac{3}{8})\\ =6\end{array}$

d.

To determine

To evaluate $\underset{x\to 2}{\lim }g(x)-f(x)$

Answer to Problem 53E

  $\frac{-125}{8}$

Explanation of Solution

Given :

  $\begin{array}{l}f(x)=x^4\\ g(x)=\frac{\sqrt{x^2+5}}{2x^2}\end{array}$

  $\begin{array}{l}\underset{x\to 2}{\lim }g(x)-f(x)\\ \underset{x\to 2}{=\lim }g(x)-\underset{x\to 2}{\lim }f(x)\\ =\frac{\sqrt{x^2+5}}{2x^2}-x^4\\ =\frac{\sqrt{2^2+5}}{2(2)}-2^4\\ =\frac{3}{8}-16\\ =\frac{3-128}{8}\\ =\frac{-125}{8}\end{array}$