Question

Each of the limits in Exercises 7–12 is of the indeterminate form $0·\infty$or $\infty ·0$. Rewrite each limit so that it is (a) in the form $\frac{0}{0}$and then (b) in the form $\frac{\infty }{\infty }$. Then (c) determine which of these indeterminate forms would be easier to work with when applying L’Hopital’s rule.

10.$\underset{x\to 0}{\lim }{x}^3\ln x$

Short answer

Part (a) The limit in the form $\frac{0}{0}$is localid="1648575908433" $\underset{x\to 0}{\lim }\frac{x^3}{{\displaystyle \frac{1}{\ln x}}}$

Part (b) The limit in the form $\frac{\infty }{\infty }$is $\underset{x\to 0}{\lim }\frac{\ln x}{{\displaystyle \frac{1}{x^3}}}$

Part (c) Part (b) is easier to work with L'Hospital's rule.

Step-by-step solution

Part (a) Step 1. Given information

We have been given the limit$\underset{x\to 0}{\lim }{x}^3\ln x$

Part (a) Step 2. Write the limit in the form 00

$$\begin{gathered} \underset{x\to 0}{\lim }{x}^3\ln x \\ =\underset{x\to 0}{\lim }\frac{x^3}{{\displaystyle \frac{1}{\ln x}}} \end{gathered}$$

Part (b) Step 1. Write the limit in the form ∞∞

$$\begin{gathered} \underset{x\to 0}{\lim }{x}^3\ln x \\ =\underset{x\to 0}{\lim }\frac{\ln x}{{\displaystyle \frac{1}{x^3}}} \end{gathered}$$

Part (c) Step 1. Determine which of these indeterminate forms would be easier to work with when applying L’Hopital’s rule.

$$\begin{gathered} \underset{x\to 0}{\lim }{x}^3\ln x \\ =\underset{x\to 0}{\lim }\frac{\ln x}{{\displaystyle \frac{1}{x^3}}} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x}}}{-3x^{-4}} \\ =\underset{x\to 0}{\lim }-\frac{1}{3}{x}^3 \\ =0 \end{gathered}$$

Therefore, part (b) is easier to work with L'Hospital's rule.