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Examine the differentiability of $f$,where $f$ is defined by $f(x)=\left \{\begin{array}{1 1}x^2\sin\large\frac{1}{x}, & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$at $x=0$

$\begin{array}{1 1} \text{Differentiable at x=0} \\\text{ Differentiable at x neq 0} \end{array} $

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Toolbox:
  • A function is not differentiable if LHL $\neq$ RHL.
  • A function is not differentiable if LHL or RHL does not exist.
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}x^2\sin\large\frac{1}{x}, & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$at $x=0$
We have left hand differentiability at $x=0$
$\qquad\qquad=\lim\limits_{\large x\to 0^-}\large\frac{f(x)-f(0)}{x-0}$
$\Rightarrow$ LHD at $x=0$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{f(0-h)-f(0)}{0-h-0}$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{(-h)^2\sin\big(\large\frac{1}{-h}\big)-0}{-h}$
$\qquad\qquad=\lim\limits_{\large h\to 0}h\sin\big(\large\frac{1}{h}\big)$
$\qquad\qquad=0\times$ (an oscillating number between -1 and 1)
$\qquad\qquad=0$
Step 2:
The right hand differentiability at $x=0$
$\qquad\qquad=\lim\limits_{\large x\to 0^+}\large\frac{f(x)-f(0)}{x-0}$
$\Rightarrow$ RHD at $x=0$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{f(0+h)-f(0)}{0+h-0}$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{f(h)-f(0)}{h}$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{h^2\sin\big(\large\frac{1}{h}\big)-0}{h}$
$\qquad\qquad=\lim\limits_{\large h\to 0}h\sin\big(\large\frac{1}{h}\big)$
$\qquad\qquad=0\times$ (an oscillating number between -1 and 1)
$\qquad\qquad=0$
Therefore LHD at $x=0$=RHD at $x=0\Rightarrow 0$
So $f(x)$ is differentiable at $x=0$
answered Jun 27, 2013 by sreemathi.v
 

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