Q. 15C - Calculus

Question

To prove that the limit forms in Theorem \(1.33\) are indeterminate, we need only list explicit examples of limits that do and do not exist for each form. Do so for each of the limit forms from Exercises \(15–21\). For the last three forms you may want to experiment with a graphing utility to find your examples.

\( \frac{0}{0}\) that approaches (a) \(0\)   (b) \(2\)   (c) \(\infty \)

Step-by-Step Solution

Answer

Part (a) \(0\)

Part (b) \(2\)

Part (c) \(\infty\)

Part a. Step 1. Given information

The given limit form is \(\frac{0}{0}\) that approaches \(0\).

Part a. Step 2. Calculation

Consider the limit expression

\(\lim_{x\rightarrow 2}\frac{(x-2)^{2}}{x-2}\)

The value of the above expression is calculated as,

\(\lim_{x\rightarrow 2}\frac{(x-2)^{2}}{x-2}=\lim_{x\rightarrow 2}\frac{(x-2)(x-2)}{x-2}\)

                           \( =\lim_{x\rightarrow 2}(x-2)\)

                           \( =2-2\)

                            \(=0\)

Part b. Step 1. Given information

The given limit form is \(\frac{0}{0}\) that approaches \(2\).

Part b. Step 2. Calculation

Consider the limit expression.

\(\lim_{x\rightarrow 2}\frac{2x-4}{x-2}\)

The value of the above expression is calculated as,

\(\lim_{x\rightarrow 2}\frac{2x-4}{x-2}=\lim_{x\rightarrow 2}\frac{2(x-2)}{x-2}\)

                            \(=\lim_{x\rightarrow 2}2\)

                            \(=2\)

Part c. Step 1. Given information

The given limit form is \(\frac{0}{0}\) that approaches \(\infty\).

Part c. Step 2. Calculation

Consider the limit expression.

\lim_{x\rightarrow 2}\frac{x-2}{(x-2)^{2}}

The value of the above expression is calculated as,

\(\lim_{x\rightarrow 2}\frac{x-2}{(x-2)^{2}}=\lim_{x\rightarrow 2}\frac{(x-2)}{(x-2)(x-2)}\)

                             \(=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\)

                             \(=\frac{1}{2-2}\)

                             \(=\infty\)