## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1162: 46

#### Answer

A function f is continuous at a number a, when it satisfies the following conditions: (a) The function f is defined at a. (b) The value of $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$

#### Work Step by Step

A function f is continuous at a number a, if it satisfies the following conditions: (a) The function f is defined at a. (b) The value of $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists. (c) $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ For example, consider a function $f\left( x \right)=x+2$ To check whether the function is continuous at the point $a=3$ or not, find the value of $f\left( x \right)$ at $a=3$, $f\left( 3 \right)=3+2=5$ The function is defined at the point $a=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,x+2$, \begin{align} & \,\underset{x\to 3}{\mathop{\lim }}\,x+2=\underset{x\to 3}{\mathop{\lim }}\,x+2 \\ & =3+2 \\ & =5 \end{align} Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,x+2=5$ From the above two steps, $\,\underset{x\to 3}{\mathop{\lim }}\,x+2=5=f\left( 3 \right)$ Thus, the function satisfies all the conditions of being continuous. Hence, the function $f\left( x \right)=x+2$ is continuous at $3$.

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