Q. 18 - 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. 

$\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}$

Step-by-Step Solution

Answer

Part (a) The value of $\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}$ using L'Hopitals rule $=0$

Part (b) The value of $\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}$ without using L'Hopital rule $=0$

Part (a) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}$

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

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}=\underset{x\to \infty }{\lim }\frac{3^{x }\ln 3}{0-4^x \ln 4} \\ =-\frac{\ln 3}{\ln 4}\underset{x\to \infty }{\lim }\frac{3^x}{4^x} \\ =-\frac{\ln 3}{\ln 4 }\underset{x\to \infty }{\lim }{\left(\frac{3}{4}\right)}^x \\ =-\frac{\ln 3}{\ln 4}.0 [\sin ce,\frac{3}{4}<1,\underset{x\to \infty }{\lim }{\left(\frac{3}{4}\right)}^x=0] \\ =0 \end{gathered}$$

Therefore,the value of the limit is $0$

Part (b) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}$

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

The limit without using L'Hopitals rule is given below 

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

Therefore,the value of $\underset{x\to \infty }{\lim }\frac{3^x}{1-4^x}=0$