Chapter 3: Q. 25 (page 310)

Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are not
$\lim_{x \to \infty} \frac{2^{- x}}{x^{2} + 1}$

Short Answer

Expert verified

The value of the given limit $\lim_{x \to \infty} \frac{2^{- x}}{x^{2} + 1} = 0$

Step by step solution

Step 1. Given information

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

Step 2. Using L’Hopital’s rule find the value of given limit

The limit using L’Hopitals rule is given below:

$\lim_{x \to \infty} \frac{2^{- x}}{x^{2} + 1} = \lim_{x \to \infty} \frac{\frac{2^{- x}}{x^{2}}}{1 + \frac{1}{x^{2}}} = \lim_{x \to \infty} \frac{\frac{1}{x^{2} 2^{x}}}{1 + \frac{1}{x^{2}}} = \frac{\frac{1}{\infty}}{1 + \frac{1}{\infty}} = \frac{0}{1} = 0$

Therfore,the value of the limit $= 0$

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Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are not
$\lim_{x \to \infty} \frac{2^{- x}}{x^{2} + 1}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Using L’Hopital’s rule find the value of given limit

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Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are notlimx→∞2-xx2+1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Using L’Hopital’s rule find the value of given limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Statistics

Theoretical and Mathematical Physics

Mechanics Maths

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are not
$\lim_{x \to \infty} \frac{2^{- x}}{x^{2} + 1}$