Chapter 14.2, Problem 38AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to -1}{\lim }\frac{x^3+x^2+3x+3}{x^4+x^3+2x+2}$

To determine

To find: ${\lim }_{x\to \text{ -1}}\frac{x^3\text{ + }{x}^2\text{ + 3}x\text{ + 3}}{x^4\text{ + }{x}^3\text{ + 2}x\text{ + 2}}$

Answer to Problem 38AYU

Solution:

4

Explanation of Solution

Given:

${\lim }_{x\to \text{ -1}}\frac{x^3\text{ + }{x}^2\text{ + 3}x\text{ + 3}}{x^4\text{ + }{x}^3\text{ + 2}x\text{ + 2}}$

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 $f(x)={\lim }_{x\to \text{ -1}}\frac{x^3\text{ + }{x}^2\text{ + 3}x\text{ + 3}}{x^4\text{ + }{x}^3\text{ + 2}x\text{ + 2}}$, at $f\left(-1\right)$, the value is $\frac{0}{0}$ which is undefined, so we have to simplify it.

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

Now the quotient formula can be used.

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

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

So ${\lim }_{x\to \text{ -1}}\frac{x^3\text{ + }{x}^2\text{ + 3}x\text{ + 1}}{x^4\text{ + }{x}^3\text{ + 2}x\text{ + 2}}$ value is 4.