Determine if f defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^2}\sin \cfrac{1}{x},}&{if}&{x \ne 0}\\{0,}&{if}&{x = 0}\end{array}} \right.$ a continuous function?}}

We have,$f(0) = 0$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} {(0 - h)^2}\sin \cfrac{1}{{(0 - h)}} = \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} \left( { - {h^2}\sin \cfrac{1}{h}} \right)$

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 + h\atop\scriptstyle h \to 0} {(0 + h)^2}\cfrac{1}{{(0 + h)}} = \mathop {\lim }\limits_{\scriptstyle x \to 0 + h\atop\scriptstyle h \to 0} {h^2}\sin \cfrac{1}{h} = 0$

therefore, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$\Rightarrow$ f is continuous at x $=$ 0.

For x $\ne$0, f(x) is a continuous at every point.

Hence we can say that f(x) is a continuous function.