Determine if $f$ defined by $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$ is a continuous function.

(A) The given function $f$ is $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
It is evident that $f$ is defined at all points of the real line.
Let $c$ be a real number.
Case $I$: If $c \neq 0$,then $f(c) = c^2 \sin \frac{1}{c}$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 \sin \frac{1}{x}) = (\lim_{x \to c} x^2) (\lim_{x \to c} \sin \frac{1}{x}) = c^2 \sin \frac{1}{c}$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all points $x \neq 0$.
Case $II$: If $c = 0$,then $f(0) = 0$.
We know that $-1 \leq \sin \frac{1}{x} \leq 1$ for $x \neq 0$.
Multiplying by $x^2$,we get $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$.
By the Squeeze Theorem,since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$,it follows that $\lim_{x \to 0} (x^2 \sin \frac{1}{x}) = 0$.
Thus,$\lim_{x \to 0} f(x) = 0 = f(0)$.
Therefore,$f$ is continuous at $x = 0$.
From the above observations,$f$ is continuous at every point of the real line. Thus,$f$ is a continuous function.