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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
A function f(x) is defined as, $​​​​​​​​​​\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{if }\text{x}\neq3\\6,&\text{if }\text{ x}=3\end{cases}$ show that f(x) is continuous that x = 3
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Answer

Given,
$​​​​​​​​​​\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{if } \text{ x}\neq3\\6 ,&\text{if }\text{ x}=3\end{cases}$
We observe 
$(\text{LHL at x }= 3)=\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{(3-\text{h})^2-9}{(3-\text{h})-3}=\lim\limits_{\text{h} \rightarrow 0}\frac{3^2-\text{h}^2-6\text{h}-9}{3-\text{h}-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+6\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}(\text{h}-6)}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}(6-\text{h})=6$
$\text{(RHL at x = 3)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}(3-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{(3+\text{h})^2-9}{3+\text{h}-3}=\lim\limits_{\text{h} \rightarrow 0}\frac{3^2+\text{h}^2+6\text{h}-9}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+6\text{h}}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}(6+\text{h})}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}(6+\text{h})=6$
Given:
$\text{f}(3)=6$
$\therefore$ $\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}(\text{x})=\text{f}(3)$
Hence, f(x) is continuous at x = 3

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