Q. 8 - 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.

8.  $\underset{x\to 0}{\lim }\left(2^x-1\right)x^{-2}$

Step-by-Step Solution

Answer

Part (a) The limit in the form $\frac{0}{0}$ is $\underset{x\to 0}{\lim }\frac{2^x-1}{x^2}$

Part (b) The limit in the form $\frac{\infty }{\infty }$ is $\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{2^x-1}}}$

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

Part (a) Step 1. Given information

We have been given the limit $\underset{x\to 0}{\lim }\left(2^x-1\right)x^{-2}$

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

$$\begin{gathered} \underset{x\to 0}{\lim }\left(2^x-1\right)x^{-2} \\ =\underset{x\to 0}{\lim }\frac{2^x-1}{x^2} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 0}{\lim }\left(2^x-1\right)x^{-2} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{2^x-1}}} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 0}{\lim }\left(2^x-1\right)x^{-2} \\ =\underset{x\to 0}{\lim }\frac{2^x-1}{x^2} \\ =\underset{x\to 0}{\lim }\frac{2^x\ln 2}{2x} \\ =\underset{x\to 0}{\lim }\frac{2^x{\left(\ln 2\right)}^2}{2} \\ =\frac{{\left(\ln 2\right)}^2}{2} \end{gathered}$$

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