Chapter 14.2, Problem 20AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to -1}{\lim }\left(8x^5-7x^3+8x^2+x-4\right)$

To determine

To find: $\underset{x\to -1}{\lim }\left(8x^5-\text{ 7}{x}^3+\text{ 8}{x}^2\text{ + }x-\text{ 4}\right)$

Answer to Problem 20AYU

$2$

Explanation of Solution

Given:

$\underset{x\to -1}{\lim }\left(8x^5-\text{ 7}{x}^3+\text{ 8}{x}^2\text{ + }x-\text{ 4}\right)$

Formula used:

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

$\underset{x\to c}{\text{lim}}\left[f\left(x\right)\text{ + }g\left(x\right)\right]=\underset{x\to c}{\text{lim}}f\left(x\right)\text{ + }\underset{x\to c}{\text{lim}}g\left(x\right)$

$\underset{x\to c}{\text{lim}}\left[f\left(x\right)\cdot g\left(x\right)\right]=\underset{x\to c}{\text{lim}}f\left(x\right)\cdot \underset{x\to c}{\text{lim}}g\left(x\right)$

Calculation:

So $\underset{x\to -1}{\lim }\left(8x^5-\text{ 7}{x}^3+\text{ 8}{x}^2\text{ + }x-\text{ 4}\right)=\underset{x\to -1}{\lim }8x^5-\underset{x\to -1}{\lim }7x^3+\underset{x\to -1}{\lim }8x^2\text{ + }\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }4$.

$\underset{x\to -1}{\text{lim}}8x^5=\underset{x\to -1}{\text{lim}}\text{8 }\times \underset{x\to -1}{\text{lim}}{x}^5\text{ = }8\times -1=-8$

$\underset{x\to -1}{\text{lim}}7x^3=7\times -1=-7$

$\underset{x\to -1}{\text{lim}}8x^2=8\times 1=8$

$\underset{x\to -1}{\text{lim}}x=-1$

$\underset{x\to -1}{\text{lim}}4=4$

$\underset{x\to -1}{\lim }\left(8x^5-\text{ 7}{x}^3+\text{ 8}{x}^2\text{ + }x-\text{ 4}\right)=-8-\left(-7\right)+8-1-4=-8+7+8-5=2$

So $\underset{x\to -1}{\lim }\left(8x^5-\text{ 7}{x}^3+\text{ 8}{x}^2\text{ + }x-\text{ 4}\right)=2$.