Q. 17C
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. For the last three forms you may want to experiment with a graphing utility to find your examples.
\(0\cdot \infty\) that approaches (a) \(0\) (b) \(1\) (c) \(\infty\)
Step-by-Step Solution
VerifiedPart a. The proof is explained below.
Part b. The proof is explained below.
Part c. The proof is explained below.
The given limit form is \(0\cdot \infty\) that approaches \(0\).
Consider the limit expression.
\(\lim_{x\rightarrow \infty}x\left ( \frac{1}{x^{2}} \right )\)
The value of the above expression is calculated as,
\(\lim_{x\rightarrow \infty}x\left ( \frac{1}{x^{2}} \right )=\lim_{x\rightarrow \infty}x\left ( \frac{1}{x} \right )\)
\(=\frac{1}{\infty}\)
\(=0\)
The given limit form is \(0\cdot \infty\) that approaches \(1\).
Consider the limit expression.
\(\lim_{x\rightarrow \infty}x\left ( \frac{1}{x} \right )\)
The value of the above expression is calculated as,
\(\lim_{x\rightarrow \infty}x\left ( \frac{1}{x} \right )=\lim_{x\rightarrow \infty} 1\)
\(=1\)
The given limit form is \(0\cdot \infty\) that approaches \(\infty\).
Consider the limit expression.
\(\lim_{x\rightarrow 1}x^{2}\left ( \frac{1}{x} \right )\)
The value of the above expression is calculated as,
\(\lim_{x\rightarrow 1}x^{2}\left ( \frac{1}{x} \right )=\lim_{x\rightarrow 1} x\)
\(\infty\)