# Chapter 11 - Mid-Chapter Check Point - Page 1165: 21

The function f\left( x \right)=\left\{ \begin{align} & \frac{{{\left( x+3 \right)}^{2}}-9}{x}\text{ if }x\ne 0 \\ & 6\text{ if }x=0 \end{align} \right. is continuous at $0$.

#### Work Step by Step

Find the value of $f\left( x \right)$ at $a=0$. From the definition of the function, for $a=0$, the function takes the value $6$. Thus, $f\left( 0 \right)=6$ The function is defined at the point $a=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ , First find the left hand limit of $\,f\left( x \right)$ , As x nears $0$ from the left, the function $\,f\left( x \right)$ takes the value $\frac{{{\left( x+3 \right)}^{2}}-9}{x}$. Thus, the value of $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is, \begin{align} & \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{\left( x+3 \right)}^{2}}-9}{x} \\ & =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x+9-9}{x} \\ & =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x}{x} \\ & =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( x+6 \right) \end{align} Taking the limit inside, \begin{align} & \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x+6 \\ & =0+6 \\ & =6 \end{align} Thus, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6$ Now find the right hand limit of $\,f\left( x \right)$. As x nears $0$ from the right, the function $\,f\left( x \right)$ takes the value $\frac{{{\left( x+3 \right)}^{2}}-9}{x}$. Thus, the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is, \begin{align} & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{\left( x+3 \right)}^{2}}-9}{x} \\ & =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x+9-9}{x} \\ & =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x}{x} \\ & =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( x+6 \right) \end{align} Taking the limit inside, \begin{align} & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x+6 \\ & =0+6 \\ & =6 \end{align} Thus, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6$ The left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6$ From the above steps, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6=f\left( 0 \right)$ Since the function satisfies all the properties of being continuous, Hence, the function f\left( x \right)=\left\{ \begin{align} & \frac{{{\left( x+3 \right)}^{2}}-9}{x}\text{ if }x\ne 0 \\ & 6\text{ if }x=0 \end{align} \right. is continuous at $0$.

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