Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given information 

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

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

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{2^{-x}}{x^2+1}=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{2^{-x}}{x^2}}}{1+{\displaystyle \frac{1}{x^2}}} \\ =\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{1}{x^2{2}^x}}}{1+{\displaystyle \frac{1}{x^2}}} \\ =\frac{{\displaystyle \frac{1}{\infty }}}{1+{\displaystyle \frac{1}{\infty }}} \\ =\frac{0}{1} \\ =0 \end{gathered}$$

Therfore,the value of the limit $=0$