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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
For what value of k is the following function continuous at x = 1
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$ 
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Answer

Given,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{\text{x} \rightarrow 1}\text{f}\text{(x)}=\text{f}(1)$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^2-1}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{(x}-1)(\text{x}+1)}{\text{x}-1}=\text{k}$
$\lim\limits_{\text{x} \rightarrow 1}(\text{x}+1)=\text{k}$
$\text{k}=2$

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