Chapter 14.1, Problem 10AYU

$\underset{x\to 0}{\lim }\frac{2-x}{x^2\text{ + 4}}$

To determine

To find: The limit of $\underset{x\to 0}{\lim }\frac{\left(2-x\right)}{\left(x^2+4\right)}$.

Answer to Problem 10AYU

Solution:

$\frac{1}{2}$

Explanation of Solution

Given:

$\underset{x\to 0}{\lim }\frac{\left(2-x\right)}{\left(x^2+4\right)}$

Calculation:

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