Chapter 12.1, Problem 44E

(a)

To determine

To find : $\underset{x\to 2}{\lim }f\left(x\right)$

Answer to Problem 44E

  $\underset{x\to 2}{\lim }f\left(x\right)=2$

Explanation of Solution

Given functions : $f\left(x\right)=\frac{x}{3-x},g\left(x\right)=\sin \pi x$

calculation:

Since the function $f\left(x\right)$ is defined at x =2, so we will evaluate the limit by using method of direct substitution.

  $\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\frac{x}{3-x}=\frac{2}{3-2}=2$

(b)

To determine

To find : $\underset{x\to 2}{\lim }g\left(x\right)$

Answer to Problem 44E

  $\underset{x\to 2}{\lim }g\left(x\right)=0$

Explanation of Solution

Given functions : $f\left(x\right)=\frac{x}{3-x},g\left(x\right)=\sin \pi x$

calculation:

Since the function $g\left(x\right)$ is defined at x =2, so we will evaluate the limit by using method of direct substitution.

  $\underset{x\to 2}{\lim }g\left(x\right)=\underset{x\to 2}{\lim }\sin \pi x=\sin 2\pi =0$

(c)

To determine

To find : $\underset{x\to 2}{\lim }\left[f\left(x\right)g\left(x\right)\right]$

Answer to Problem 44E

  $\underset{x\to 2}{\lim }\left[f\left(x\right)g\left(x\right)\right]=0$

Explanation of Solution

Given functions : $f\left(x\right)=\frac{x}{3-x},g\left(x\right)=\sin \pi x$

calculation:

From previous parts, we have

  $\underset{x\to 2}{\lim }f\left(x\right)=2,\underset{x\to 2}{\lim }g\left(x\right)=0$

From algebra of limits,

  $\underset{x\to 2}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to 2}{\lim }f\left(x\right).\underset{x\to 2}{\lim }g\left(x\right)=2\times 0=0$

(d)

To determine

To find : $\underset{x\to 2}{\lim }\left[g\left(x\right)-f\left(x\right)\right]$

Answer to Problem 44E

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

Explanation of Solution

Given functions : $f\left(x\right)=\frac{x}{3-x},g\left(x\right)=\sin \pi x$

calculation:

From previous parts, we have

  $\underset{x\to 2}{\lim }f\left(x\right)=8,\underset{x\to 2}{\lim }g\left(x\right)=\frac{3}{8}$

From algebra of limits,

  $\underset{x\to 2}{\lim }\left[g\left(x\right)-f\left(x\right)\right]=\underset{x\to 2}{\lim }g\left(x\right)-\underset{x\to 2}{\lim }f\left(x\right)=0-2=-2$