Chapter 11.3, Problem 14E

Solution

a)

To explain: Why you cannot use substitution to find the limit $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ .

When one substitutes -1 for $x$ then the denominator becomes zero, so one cannot use substitution to find the limit $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ .

Given information: The given limit is $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ .

Calculation:

Substitute $2$ for $x$ in the given limit then the denominator becomes zero as shown below:

  $\begin{array}{c}x-2=2-2\\ =0\end{array}$

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

b)

To find: The limit $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ algebraically if it exists.

The limit $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ exists and is found by factoring numerator as $5$ .

Given information: The given limit is $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}$ .

Calculation:

Simplify the expression by factoring the numerator and diving out the common factors and then evaluate the limit.

  $\begin{array}{c}\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}=\underset{x\to 2}{\lim }\frac{x^2\left(x-2\right)+\left(x-2\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x^2+1\right)}{x-2}\\ ={\left(2\right)}^2+1\\ =5\end{array}$

Thus, the given limit exists and is found as $\underset{x\to 2}{\lim }\frac{x^3-2x^2+x-2}{x-2}=5$ .