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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{{1}-\text{x}^\text{n}}{1-\text{x}}, & \text{x} \neq1\\\text{n}-1, & \text{ x} = 1\end{cases}\text{ n }\in\ \text{N at x}=1$
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Answer

We want, to check the continuity at x = 1
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1-\text{h})^\text{n}}{1-(1-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}-\frac{\text{n(n-1)}}{2!}\text{h}+\dots$
$=\text{n}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 1^+}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}\text{(1+h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1+\text{h})^\text{n}}{1-(1+\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}+\frac{\text{n}(\text{n}-1)}{2!}\text{h}+\dots$
$=\text{n}$
$\text{f}(1)=\text{n}-1$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}( 1)$
Hence, funtion is discontinuous at x = 1
This is removable discotinuity.

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