Q. 59 - Calculus

Question

Calculate each limit in Exercises 35–80.  

$\underset{x\to -\infty }{lim}\frac{3^x-5^x}{4^x}$

Step-by-Step Solution

Answer

The limit is infinity

Step 1. Given information

The given expression is $\underset{x\to -\infty }{lim}\frac{3^x-5^x}{4^x}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to -\infty }{lim}\frac{3^x-5^x}{4^x}==\underset{x\to -\infty }{lim}\left(\frac{3^x-5^x}{4^x}\times \frac{{\displaystyle \frac{1}{4^x}}}{{\displaystyle \frac{1}{4^x}}}\right) \\ =\underset{x\to -\infty }{lim}\left(\frac{{\left({\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^x-{\left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^x}{1}\right) \\ =\left(\frac{{\left({\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^{-\infty }-{\left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^{-\infty }}{1}\right) \\ =\frac{\infty -\infty }{1} \\ =\infty \end{gathered}$$