Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1160: 5


Yes, the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$, First check whether the function is defined at the point $ a $ or not. Find the value of $ f\left( x \right)$ at $ a=3$, $\begin{align} & f\left( 3 \right)=\frac{{{\left( 3 \right)}^{2}}+4}{\left( 3 \right)-2} \\ & =\frac{9+4}{3-2} \\ & =13 \end{align}$ The function is defined at the point $ a=3$. Now find the value of $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}$, $\begin{align} & \,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2} \\ & =\frac{\,\underset{x\to 3}{\mathop{\lim }}\,\left( {{x}^{2}}+4 \right)}{\,\underset{x\to 3}{\mathop{\lim }}\,\left( x-2 \right)}\, \\ & =\frac{{{\left( 3 \right)}^{2}}+4}{3-2} \\ & =13 \end{align}$ Thus, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}+4}{x-2}=13\text{ and }f\left( 3 \right)=13$ Therefore, $\,\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=f\left( 3 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\frac{{{x}^{2}}+4}{x-2}$ is continuous at $3$.
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