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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability3 Marks
Question
If function $f(x)=\left\{\begin{array}{cc}x^5 \sin \frac{1}{x}, & x \neq 0 \\ k & x=0\end{array}\right.$, is continuous at $x =0$, find the value of $k$.
✓

Answer

Function at $x=0$
$
\begin{aligned}
f(x) & =k, \quad \text { when } x=0 \\
\therefore f(0) & =k
\end{aligned}
$
value of R.H.L.$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}(0+h)^5 \sin \frac{1}{0+h} \\
& =\lim _{h \rightarrow 0} h^5 \sin \frac{1}{h}=\lim _{h \rightarrow 0} h^5 \times \lim _{h \rightarrow 0} \sin \frac{1}{h} \\
& =0 \times\{\text { any constant between }-1 \text { and } 1\} \\
& =0
\end{aligned}
$
is continuous at $x=0$
$
\begin{aligned}
\therefore & & f(0) & =\lim _{h \rightarrow 0} f(0+h) \\
\Rightarrow & & k & =0 \text {}
\end{aligned}
$

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