Q. 58 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to \infty }{lim}\frac{2^x-4^{-x}}{3^x}$

Step-by-Step Solution

Answer

The limit is 0

Step 1. Given information

The given expression is $\underset{x\to \infty }{lim}\frac{2^x-4^{-x}}{3^x}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{2^x-4^{-x}}{3^x}=\underset{x\to \infty }{lim}\left(\frac{2^x-{\displaystyle \frac{1}{4^x}}}{3^x}\times \frac{{\displaystyle \frac{1}{3^x}}}{{\displaystyle \frac{1}{3^x}}}\right) \\ =\underset{x\to \infty }{lim}\left(\frac{{\left({\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^x-{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}\right)}^x}{1}\right) \\ =\left(\frac{{\left({\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\infty }-{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}\right)}^{\infty }}{1}\right) \\ =\frac{0-0}{1} \\ =0 \end{gathered}$$