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Q. Assertion : $f(x)=x^{n} \sin \left(\frac{1}{x}\right)$ is differentiable for all real values of $x(n \geq 2)$.
Reason : For $n \geq 2, \displaystyle\lim _{x \rightarrow 0} f(x)=0$

Continuity and Differentiability

Solution:

$\displaystyle\lim _{x \rightarrow 0} f(x)=\displaystyle\lim _{x \rightarrow 0} x^{n} \sin \left(\frac{1}{x}\right)=0$ for positive integer $n$
Now $f(0)$ does not exist, hence function is not continuous at $x=0$