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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1},& \text{for }\text{ x}\neq1 \\2,&\text{for }\text{ x}=1\end{cases}$ Find whether f (x) is continuous at x = 1.
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Answer

Given,
$\text{f}\text{(x)}=\frac{\text{x}^2-1}{\text{x}-1},\text{ if}\text{ x}\neq1$
$\text{f}\text{(x)}=2,\text{ if}\text{ x}=1$
We observe
$\text{(LHL at x = 1)}$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{ (x)}=\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})=\lim\limits_{\text{x} \rightarrow 0}\frac{(1-\text{h})^2-1}{(1-\text{h})^2-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\text{h}^2-2\text{h}-1}{1-\text{h}-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{h}^2-2\text{h}}{-\text{h}}$
$\lim\limits_{\text{x} \rightarrow 0}2-\text{h}$
$= 2$
$(\text{RHL at x}=1)$
$\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}(1+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1+\text{h})^2-1}{(1+\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{1+\text{h}^2+2\text{h}-1}{1+\text{h}-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+2\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}\text{h}+2$
$=2$
Also f(x) = 2
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\text{f}(1)$
Hence f(x) is continuous at x = 1.

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