# Determine if f defined by

Question:

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?

Solution:

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 $f(c)=c^{2} \sin \frac{1}{c}$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\left(\lim _{x \rightarrow c} x^{2}\right)\left(\lim _{x \rightarrow c} \sin \frac{1}{x}\right)=c^{2} \sin \frac{1}{c}$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Thereforef is continuous at all points  0

Case II:

If $c=0$, then $f(0)=0$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)$

It is known that, $-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0$

$\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2}$

$\Rightarrow \lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2}$

$\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0$

$\Rightarrow \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=0$

Similarly, $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)$

Thereforef is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.