Chapter 2.1, Problem 6E

To determine

To find: The value of $\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)$ with the help of properties of limits.

Answer to Problem 6E

The value of $\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)$ is $\frac{c^4-c^3+1}{c^2+9}$ .

Explanation of Solution

Given information:

The limit is $\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)$ .

Calculation:

Use the quotient rule for limit.

  $\underset{x\to c}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\underset{x\to c}{\lim }f\left(x\right)}{\underset{x\to c}{\lim }g\left(x\right)}$

Simplify the given limit.

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

Use the sum and difference rule for the limits.

  $\underset{x\to c}{\lim }\left(f\left(x\right)\pm g\left(x\right)\right)=\underset{x\to c}{\lim }f\left(x\right)\pm \underset{x\to c}{\lim }g\left(x\right)$

Simplify the limit.

  $\begin{array}{c}\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)=\frac{\underset{x\to c}{\lim }\left(x^4-x^3+1\right)}{\underset{x\to c}{\lim }\left(x^2+9\right)}\\ =\frac{\underset{x\to c}{\lim }{x}^4-\underset{x\to c}{\lim }{x}^3+\underset{x\to c}{\lim }1}{\underset{x\to c}{\lim }{x}^2+\underset{x\to c}{\lim }9}\end{array}$

Use the constant rule and power rule for limit.

  $\begin{array}{c}\underset{x\to c}{\lim }\left(a\right)=a\\ \underset{x\to c}{\lim }\left(x^n\right)=c^n\end{array}$

Simplify the limit.

  $\begin{array}{c}\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)=\frac{\underset{x\to c}{\lim }{x}^4-\underset{x\to c}{\lim }{x}^3+\underset{x\to c}{\lim }1}{\underset{x\to c}{\lim }{x}^2+\underset{x\to c}{\lim }9}\\ =\frac{c^4-c^3+1}{c^2+9}\end{array}$

Therefore, the value of $\underset{x\to c}{\lim }\left(\frac{x^4-x^3+1}{x^2+9}\right)$ is $\frac{c^4-c^3+1}{c^2+9}$ .