Q. 12 - Calculus

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.

12.  $\underset{x\to 1}{\lim }\sqrt{x-1}\ln \left(x-1\right)$

Step-by-Step Solution

Answer

Part (a) The limit in the form $\frac{0}{0}$ is $\underset{x\to 1}{\lim }\frac{\sqrt{x-1}}{{\displaystyle \frac{1}{\ln \left(x-1\right)}}}$

Part (b) The limit in the form $\frac{\infty }{\infty }$ is $\underset{x\to 1}{\lim }\frac{\ln \left(x-1\right)}{{\displaystyle \frac{1}{\sqrt{x-1}}}}$

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

Part (a) Step 1. Given information

We have been given the limit $\underset{x\to 1}{\lim }\sqrt{x-1}\ln \left(x-1\right)$

Part (a) Step 2. Write the limit in the form 0 0

$$\begin{gathered} \underset{x\to 1}{\lim }\sqrt{x-1}\ln \left(x-1\right) \\ =\underset{x\to 1}{\lim }\frac{\sqrt{x-1}}{{\displaystyle \frac{1}{\ln \left(x-1\right)}}} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 1}{\lim }\sqrt{x-1}\ln \left(x-1\right) \\ =\underset{x\to 1}{\lim }\frac{\ln \left(x-1\right)}{{\displaystyle \frac{1}{\sqrt{x-1}}}} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 1}{\lim }\sqrt{x-1}\ln \left(x-1\right) \\ =\underset{x\to 1}{\lim }\frac{\ln \left(x-1\right)}{{\displaystyle \frac{1}{\sqrt{x-1}}}} \\ =\underset{x\to 1}{\lim }\frac{{\displaystyle \frac{1}{x-1}}}{-{\displaystyle \frac{1}{2}}{\left(x-1\right)}^{-{\displaystyle \frac{3}{2}}}} \\ =\infty \end{gathered}$$

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