Q. 61 - Calculus

Question

Calculate each limit in Exercises 35–80.  

$\underset{x\to \infty }{lim}\frac{2e^{1.5x}}{3e^{2x}+e^{1.5x}}$

Step-by-Step Solution

Answer

The limit is 0

Step 1. Given information

The given expression is $\underset{x\to \infty }{lim}\frac{2e^{1.5x}}{3e^{2x}+e^{1.5x}}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{2e^{1.5x}}{3e^{2x}+e^{1.5x}}=\underset{x\to \infty }{lim}\frac{2e^{1.5x}}{3e^{2x}+e^{1.5x}}\times \frac{{\displaystyle \frac{1}{e^{1.5x}}}}{{\displaystyle \frac{1}{e^{1.5x}}}} \\ =\underset{x\to \infty }{lim}\frac{2}{3e^{2x-1.5x}+1} \\ =\underset{x\to \infty }{lim}\frac{2}{3e^{0.5x}+1} \\ =\frac{2}{3e^{0.5(\infty )}+1} \\ =\frac{2}{3(\infty )+1} \\ =0 \end{gathered}$$