Chapter 11.3, Problem 13E

Solution

a)

To explain: Why you cannot use substitution to find the limit $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ .

When one substitutes -1 for $x$ then the denominator becomes zero, so one cannot use substitution to find the limit $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ .

Given information: The given limit is $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ .

Calculation:

Substitute $-1$ for $x$ in the given limit then the denominator becomes zero as shown below:

  $\begin{array}{c}x+1=\left(-1\right)+1\\ =0\end{array}$

Thus, the denominator becomes zero so one cannot use substitution to find the given limit.

b)

To find: The limit $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ algebraically if it exists.

The limit $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ exists and is found by factoring numerator and denominator as $3$ .

Given information: The given limit is $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}$ .

Calculation:

Simplify the expression by factoring the numerator using sum of cubes $x^3+1=\left(x+1\right)\left(x^2-x+1\right)$ and diving out the common factors and then evaluate the limit.

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

Thus, the given limit exists and is found as $\underset{x\to -1}{\lim }\frac{x^3+1}{x+1}=3$ .