If f(x) = l x^2 1 x , ; ; ; when , , x 0 , , , , , , , , , , , , … - Vidyadip

MCQ

If $f(x) = \left\{ \begin{array}{l}{x^2}\sin \frac{1}{x},\;\;\;{\rm{when\,\, }}x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,\,\,\,{\rm{when\,\,}}\,x = 0\end{array} \right.$, then

A
$f(0 + 0) = 1$
B
$f(0 - 0) = 1$
C
$f$ is continuous at $x = 0$
D
None of these

✓

Answer

$\mathop {\lim }\limits_{x \to {0^ + }} \,f(x) = {x^2}\sin \frac{1}{x},$ but $ - 1 \le \sin \frac{1}{x} \le 1$ and $x \to 0$
Therefore, $\mathop {\lim }\limits_{x \to {0^ + }} \,f(x) = 0 = \mathop {\lim }\limits_{x \to {0^ - }} \,f(x) = f(0)$
Hence $f(x)$ is continuous at $x = 0.$

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