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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} $

1 Answer

  • 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$
$f$ is continuous for all $x\in R$
answered May 29, 2013 by sreemathi.v
edited May 29, 2013 by sreemathi.v

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