Chapter 14.2, Problem 19AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)$

To determine

To find: $\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)$

Answer to Problem 19AYU

$-1$

Explanation of Solution

Given:

$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)$

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 $\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)=\underset{x\to \text{ 1}}{\lim }5x^4-\underset{x\to \text{ 1}}{\lim }3x^2+\underset{x\to \text{ 1}}{\lim }6x-\underset{x\to \text{ 1}}{\lim }9$.

$\underset{x\to 1}{\text{lim}}5x^4=\underset{x\to \text{ 1}}{\text{lim}}5\times \underset{x\to \text{ 1}}{\text{lim}}{x}^4$

$=\text{ 5 }\times \text{ 1 = 5}$

$\underset{x\to \text{ 1}}{\text{lim}}3x^2=3\times \text{ 1 = 3}$

$\underset{x\to \text{ 1}}{\text{lim}}6x=6\times 1=6$

$\underset{x\to \text{ 1}}{\text{lim}}9=9$

$\underset{x\to \text{ 1}}{\lim }5x^4-\underset{x\to \text{ 1}}{\lim }3x^2+\underset{x\to \text{ 1}}{\lim }6x-\underset{x\to \text{ 1}}{\lim }9=5-3+6-9=-1$

So $\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)\text{ = }-1$.