MCQ
If $f(x) = \left\{ \begin{array}{l}\frac{1}{x}\sin {x^2},\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,x = 0\end{array} \right.$, then
  • A
    $\mathop {\lim }\limits_{x \to 0 + } f(x) \ne 0$
  • B
    $\mathop {\lim }\limits_{x \to 0 - } f(x) \ne 0$
  • $f(x),$ is continuous at $x = 0 $
  • D
    None of these

Answer

Correct option: C.
$f(x),$ is continuous at $x = 0 $
c
(c) $f(0) = 0,\,\mathop {\lim }\limits_{x \to 0 + } f(x) = \mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0} \,x\,\left[ {\frac{{\sin {x^2}}}{{{x^2}}}} \right] = 0$.

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