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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
A function f(x) is defined as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-\text{x}-6}{\text{x}-3}&; &\text{if} \text{x}\neq3\\5 &;&\text{if}\text{ x}=3\end{cases}$
show that f(x) is continuous that x = 3.
✓

Answer

We have, to check the continuity at x = 3.
$\text{L.H.L}=\lim\limits_{\text{x} \rightarrow 3^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\frac{(3-\text{h})^2-(3-\text{h})-6}{(3-\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2-5\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}+5=5$
$\text{R.H.L}=\lim\limits_{\text{x} \rightarrow 3^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(3\text{+h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(3+\text{h})^2-(3+\text{h})-6}{(3+\text{h})-3}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+5\text{h}}{\text{h}}=\lim\limits_{\text{h}\rightarrow 0}\text{h}+5=5$
$\text{f}(3)=5$
Thus, We have, LHL = RHL = f(3) = 5
So,The function is continus at x = 3

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