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If $f(x)=x^2\sin \frac{1}{x}$,when $x\neq0$,then the value of the function f at $x=0$,so that the function is continuous at $x=0$,is

\begin{array}{1 1}(A)\;0 & (B)\;-1\\(C)\;1 & (D)\;none\;of\;these\end{array}

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  • $\lim\limits_{\large x\to 0}\large\frac{\sin x}{x}$$=1$
Step 1:
This can be written as $x\bigg(\large\frac{\sin\large\frac{1}{x}}{\Large\frac{1}{x}}\bigg)$
Applying the limits we get,
$\quad=\lim\limits_{\large x\to 0}x\bigg(\large\frac{\sin\large\frac{1}{x}}{\Large\frac{1}{x}}\bigg)$
$\lim\limits_{\large x\to 0}\large\frac{\sin\theta}{\theta}$$=1$
$\quad=0\times 1$
Hence the correct option is $A$
answered Jul 4, 2013 by sreemathi.v

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