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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find which of the function:
$\text{f(x)}=\begin{cases}|\text{x}-\text{a}|\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$
at x = a
✓

Answer

We have, $\text{f(x)}=\begin{cases}|\text{x}-\text{a}|\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$ at x = a.
At x = a $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow\text{a}^-}|\text{x}-\text{a}|\sin\frac{1}{\text{x}-\text{a}}$
$=\lim\limits_{\text{h}\rightarrow0}|\text{a}-\text{h}-\text{a}|\sin\Big(\frac{1}{\text{a}-\text{h}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}-\text{h}\sin\Big(\frac{1}{\text{h}}\Big)\ [\because\sin(-\theta)=-\sin\theta]$
= 0 × [an oscillating number between -1 and 1] = 0
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow\text{a}^+}|\text{x}-\text{a}|\sin\Big(\frac{1}{\text{x}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}|\text{a}+\text{h}-\text{a}|\sin\Big(\frac{1}{\text{a}+\text{h}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\sin\frac{1}{\text{h}}$
= 0 × [an oscillating number between -1 and 1] = 0
and f(a) = 0
$\therefore$ L.H.L = R.H.L = f(a)
So, f(x) is continuous at x = a.

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