Chapter 14.2, Problem 42AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$

To determine

To find: ${\lim }_{x\to \text{ 3}}\frac{x^3-\text{ 3}{x}^2\text{ + 4}x-\text{ 12}}{x^4-3x^3\text{ + }x-\text{ 3}}$

Answer to Problem 42AYU

Solution:

$\frac{13}{10}$

Explanation of Solution

Given:

${\lim }_{x\to \text{ 3}}\frac{x^3-\text{ 3}{x}^2\text{ + 4}x-\text{ 12}}{x^4-3x^3\text{ + }x-\text{ 3}}$

Formula used:

${\lim }_{x\to c}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to c}{ \text{lim}}f\left(x\right)}{\underset{x\to c}{ \text{lim}}g\left(x\right)}$, provided that $\underset{x\to c}{\text{lim}}g\left(x\right)\ne 0$

Calculation:

For $f\left(x\right)={\lim }_{x\to \text{ 3}}\frac{x^3-\text{ 3}{x}^2\text{ + 4}x-\text{ 12}}{x^4-3x^3\text{ + }x-\text{ 3}}$, at $f\left(-3\right)$ the value is $\frac{0}{0}$ which is undefined, so we have to simplify it.

So $\frac{x^3-\text{ 3}{x}^2\text{ + 4}x-\text{ 12}}{x^4-3x^3\text{ + }x-\text{ 3}}=\frac{\left(x^2(x-3\right)+4 \left(x-3\right)}{x^3\text{(}x-\text{ 3) + (}x-\text{ 3)}}=\frac{\left(x-3\right)\left(x^2+4\right)}{\left(x-3\right)\left(x^3+1\right)}=\frac{\left(x^2+4\right)}{\left(x^3+1\right)}$

Now the quotient formula can be used.

${\lim }_{x\to c}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to c}{\text{lim}}f\left(x\right)}{\underset{x\to c}{\text{lim}}g\left(x\right)}$, provided that $\underset{x\to c}{\text{lim}}g\left(x\right)\ne 0$

Now ${\lim }_{x\to \text{ 3}}\frac{x^2\text{ + }4}{x^3\text{ + 1}}=\frac{\underset{x\to 3}{\text{lim}}\left(x^2+4\right)}{\underset{x\to 3}{\text{lim}}\left(x^3+1\right)}=\frac{13}{10}$

So ${\lim }_{x\to \text{ 3}}\frac{x^3\text{ – 3}{x}^2\text{ + 4}x\text{ – 12}}{x^4\text{ + 3}{x}^3\text{ + }x-\text{ 3}}\frac{13}{10}$ value is $\frac{13}{10}$.