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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin(\frac{1}{\text{x}-\text{a}}), &\text{for} \text{ x} \neq\text{a}\\0,&\text{for} \text{ x} = \text{a}\end{cases}\text{ at x}=0$
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Answer

$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin\Big(\frac{1}{\text{x}-\text{a}}\Big),&\text{for }\text{x }\neq \text{a}\\0, &\text{for x } = \text{a}\end{cases}$ $(\text{LHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}-\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}-\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}-\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{-\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
= 0 $(\text{RHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}+\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}+\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}+\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
Thus, we obtian $\lim\limits_{\text{x} \rightarrow\text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow\text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$ $\therefore$ f(x) is continuous at x = a.

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