Q. 7 - 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.

7. $\underset{x\to \infty }{\lim }{2}^{-x}x$

Step-by-Step Solution

Answer

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

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

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 $\underset{x\to \infty }{\lim }{2}^{-x}x$

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

$$\begin{gathered} \underset{x\to \infty }{\lim }{2}^{-x}x \\ =\underset{x\to \infty }{\lim }\frac{2^{-x}}{{\displaystyle \frac{1}{x}}} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to \infty }{\lim }{2}^{-x}x \\ =\underset{x\to \infty }{\lim }\frac{x}{{\displaystyle 2^x}} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to \infty }{\lim }{2}^{-x}x \\ =\underset{x\to \infty }{\lim }\frac{x}{{\displaystyle 2^x}} \left[in \frac{\infty }{\infty } form\right] \\ =\underset{x\to \infty }{\lim }\frac{1}{2^x\ln 2} \\ =0 \end{gathered}$$

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