Q. 19 - Calculus

Question

Calculate each of the limits in Exercises 15–20

(a) using L’Hopital’s rule and (b) without using L’Hopital’s rule.

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

Step-by-Step Solution

Answer

Part (a) The value of $\underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}$ using L'Hopitals rule$-\infty$

Part (b) The value of $\underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}$ without using L'Hopitals rule$-\infty$

Part (a) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}$

Part (a) Step 2. Finding the value of the given limit using L’Hopital’s rule

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}=\underset{x\to \infty }{\lim }\frac{3.e^{3x}}{-2.e^{2x}} \\ =-\frac{3}{2}\underset{x\to \infty }{\lim }{e}^{3x-2x} \\ =-\frac{3}{2}\underset{x\to \infty }{\lim }{e}^x \\ =-\infty \end{gathered}$$

Therefore,the value of limit is $-\infty$

Part (b) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}$

Part (b) Step 2. Finding the value of the given limit without using L’Hopital’s rule

The limit without using L'Hopitals rule is given below 

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^{3x}}{1-e^{2x}}=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{e^{3x}}{e^{2x}}}}{{\displaystyle \frac{1}{e^{2x}}-1}} \\ =\underset{x\to \infty }{\lim }\frac{e^{3x-2x}}{{\displaystyle \frac{1}{e^{2x}}-1}} \\ =\underset{x\to \infty }{\lim }\frac{e^x}{e^{-2x}-1} \\ =-\infty \end{gathered}$$

Therefore,the value of the limit $-\infty$