# Determine if $$f$$ defined by $f(x) = \left\{ \begin{array} {1 1} x^2sin \large\frac{ 1 }{x},\normalsize & \quad\text{ if x $$\neq$$ 0 }\\ 0 ,& \quad \text{if$x$=0}\\ \end{array} \right.$ is a continuous function?

$\begin{array}{1 1} \text{f is not continuous.} \\ \text{f is continuous for all} \; x\in R \end{array}$

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL=$\lim\limits_{\large x\to 0^-}\big(x^2\sin\large\frac{1}{x}\big)$
$\quad\;\;=\lim\limits_{\large h\to 0}(-h)^2\sin\large\frac{1}{-h}$
$\quad\;\;=-\lim\limits_{\large h\to 0}h^2(\sin\large\frac{1}{h})$
$\sin\large\frac{1}{h}$ lies between $-1$ and $1$ a finite quantity.
$h^2\sin\large\frac{1}{h}$$\rightarrow 0 as h\to 0 Therefore LHL=0. Step 2: Similarly RHL= \lim\limits_{\large x\to 0^+}(x^2\sin\large\frac{1}{x})$$=0$
$f(0)=0$
LHL=RHL=f(0)
$f$ is continuous for all $x\in R$
answered May 29, 2013
edited May 29, 2013