Chapter 3: Q. 20 (page 310)

Calculate each of the limits in Exercises 15–20
(a) using L’Hopital’s rule and (b) without using L’Hopital’s rule.
$\lim_{x \to \infty} \frac{2^{x} - 1}{4^{x} - 1}$

Short Answer

Expert verified

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

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

Step by step solution

Part (a) Step 1. Given information

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

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:

$\lim_{x \to 0} \frac{2^{x} - 1}{4^{x} - 1} = \lim_{x \to 0} \frac{2^{x} \ln 2}{4^{x} \ln 4} = \frac{\ln 2}{\ln 4} \lim_{x \to 0} \frac{2^{x}}{4^{x}} = \frac{\ln 2}{\ln 4} \lim_{x \to 0} \frac{2^{x}}{2^{2 x}} = \frac{\ln 2}{\ln 4} \lim_{x \to 0} 2^{x - 2 x} = \frac{\ln 2}{\ln 4} \lim_{x \to 0} 2^{- x} = \frac{\ln 2}{\ln 2^{2}} . 1 = \frac{\ln 2}{2 \ln 2} = \frac{1}{2}$

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

Part (b) Step 1. Given information

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

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

$\lim_{x \to 0} \frac{2^{x} - 1}{4^{x} - 1} = \lim_{x \to 0} \frac{2^{x} - 1}{{(2^{x})}^{2} - 1^{2}} = \lim_{x \to 0} \frac{2^{x} - 1}{(2^{x} + 1) (2^{x} - 1)} = \lim_{x \to 0} \frac{1}{2^{x} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

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

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Calculate each of the limits in Exercises 15–20
(a) using L’Hopital’s rule and (b) without using L’Hopital’s rule.
$\lim_{x \to \infty} \frac{2^{x} - 1}{4^{x} - 1}$

Short Answer

Step by step solution

Part (a) Step 1. Given information

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

Part (b) Step 1. Given information

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

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Calculate each of the limits in Exercises 15–20(a) using L’Hopital’s rule and (b) without using L’Hopital’s rule.limx→∞2x-14x-1

Short Answer

Step by step solution

Part (a) Step 1. Given information

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

Part (b) Step 1. Given information

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

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Calculate each of the limits in Exercises 15–20
(a) using L’Hopital’s rule and (b) without using L’Hopital’s rule.
$\lim_{x \to \infty} \frac{2^{x} - 1}{4^{x} - 1}$