Q. 60 - Calculus

Question

Calculate each limit in Exercises 35–80.  

$\underset{x\to \infty }{lim}\frac{4\left(3^x\right)}{2+3^x}$

Step-by-Step Solution

Answer

The limit is 4

Step 1. Given information

The given expression is $\underset{x\to \infty }{lim}\frac{4\left(3^x\right)}{2+3^x}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{4\left(3^x\right)}{2+3^x}=\underset{x\to \infty }{lim}\frac{4\left(3^x\right)}{2+3^x}\times \frac{{\displaystyle \frac{1}{3^x}}}{{\displaystyle \frac{1}{3^x}}} \\ =\underset{x\to \infty }{lim}\frac{4}{{\displaystyle \frac{2}{3^x}}+1} \\ =\frac{4}{{\displaystyle \frac{2}{3^{\infty }}+1}} \\ =\frac{4}{{\displaystyle \frac{2}{\infty }}+1} \\ =\frac{4}{0+1} \\ =4 \end{gathered}$$