Q. 62 - Calculus

Question

Calculate each limit in Exercises 35–80.   

$\underset{x\to \infty }{lim}\frac{1-5e^{2x}}{3e^x+4e^{2x}}$

Step-by-Step Solution

Answer

The limit is $-\frac{5}{4}$

Step 1. Given information

The given expression is $\underset{x\to \infty }{lim}\frac{1-5e^{2x}}{3e^x+4e^{2x}}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{1-5e^{2x}}{3e^x+4e^{2x}}=\underset{x\to \infty }{lim}\frac{1-5e^{2x}}{3e^x+4e^{2x}}\times \frac{{\displaystyle \frac{1}{e^{2x}}}}{{\displaystyle \frac{1}{e^{2x}}}} \\ =\underset{x\to \infty }{lim}\frac{e^{-2x}-5}{3e^{-x}+4} \\ =\frac{e^{-2(\infty )}-5}{3e^{-(\infty )}+4} \\ =\frac{0-5}{3\left(0\right)+4} \\ =-\frac{5}{4} \end{gathered}$$