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Examine the continuity of the function $ f(x) = \left\{ \begin{array}{l l}\large\frac{| sin\: x |}{x}, & \quad if \;x \; \neq\; 0 \\ 1, & \quad if \;x = 0 \end{array} \right. $

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  • 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:
$f(x)=\left\{\begin{array}{1 1}\large\frac{\mid\sin x\mid}{x}&if\;x\neq 0\\1&if\;x=0\end{array}\right.$
We have LHL at $x=0$
$\Rightarrow \lim\limits_{x\to 0^-}f(x)=\lim\limits_{h\to 0}(0-h)$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{\sin(-h)}{-h}=\frac{\sin h}{-h}$
$\Rightarrow -1\lim\limits_{h\to 0}\large\frac{\sin h}{h}$
But $\large\frac{\sin h}{h}$$=1$
$\therefore -1\times \lim\limits_{h\to 0}\large\frac{\sin h}{h}$$=-1$
Step 2:
RHL at $x=0$
$\lim\limits_{x\to 0^+}f(x)=\lim\limits_f(0+h)=\lim\limits_{h\to 0}f(h)$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{\mid \sin h\mid}{h}=\lim\limits_{h\to 0}\large\frac{\sin h}{h}$$=1$
Hence we have $\lim\limits_{x\to 0^-}f(x)\neq \lim\limits_{x\to 0^+}f(x)$
Hence $f(x)$ is not continuous at the origin.
answered Sep 30, 2013 by sreemathi.v

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