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If $f(x)=\left\{\begin{array}{rr}x^2 \sin \frac{1}{x}, & \text { when } x \neq 0 \\ 0, & \text { when } x=0\end{array}\right.$ then
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The correct answer is:
$f$ is continuous at $x=0$
$\lim _{x \rightarrow 0^{+}} f(x)=x^2 \sin \frac{1}{x}$, but $-1 \leq \sin \frac{1}{x} \leq 1$ and $x \rightarrow 0$
Therefore, $\lim _{x \rightarrow 0^{+}} f(x)=0=\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
Hence $f(x)$ is continuous at $x=0$.
Therefore, $\lim _{x \rightarrow 0^{+}} f(x)=0=\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
Hence $f(x)$ is continuous at $x=0$.
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