Q. 20 - 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.

$\underset{x\to \infty }{\lim }\frac{2^x-1}{4^x-1}$

Step-by-Step Solution

Answer

Part (a) The value of $\underset{x\to \infty }{\lim }\frac{2^x-1}{4^x-1}$ using L'Hopitals rule $=\frac{1}{2}$

Part (b) The value of $\underset{x\to \infty }{\lim }\frac{2^x-1}{4^x-1}$ without using L'Hopitals rule$=\frac{1}{2}$

Part (a) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{2^x-1}{4^x-1}$

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:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{2^x-1}{4^x-1}=\underset{x\to 0}{\lim }\frac{2^x\ln 2}{4^x\ln 4} \\ =\frac{\ln 2}{\ln 4}\underset{x\to 0}{\lim }\frac{2^x}{4^x} \\ =\frac{\ln 2}{\ln 4}\underset{x\to 0}{\lim }\frac{2^x}{2^{2x}} \\ =\frac{\ln 2}{\ln 4}\underset{x\to 0}{\lim }{2}^{x-2x} \\ =\frac{\ln 2}{\ln 4}\underset{x\to 0}{\lim }{2}^{-x} \\ =\frac{\ln 2}{\ln 2^2}.1 \\ =\frac{\ln 2}{2 \ln 2} \\ =\frac{1}{2} \end{gathered}$$

Therefore,the value of limit $\frac{1}{2}$

Part (b) Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{2^x-1}{4^x-1}$

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 

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

Therefore,the value of the limit $=\frac{1}{2}$