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The function $f(x)=\large\frac{|x|}{x^2+2x}\qquad$$ x\neq 0$ and $f(0)=0$ is not continuous at $x=0$ because

$\begin{array}{1 1}(a)\;\lim\limits_{x\to 0}f(x)\neq f(0)&(b)\;\lim\limits_{x\to 0^+}f(x)\;\text{does not exist}\\(c)\;\lim\limits_{x\to 0^-}f(x)\;\text{does not exist}&(d)\;\lim\limits_{x\to 0}f(x)\;\text{does not exist}\end{array}$

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$\lim\limits_{x\to 0^+}\large\frac{|x|}{x^2+2x}$
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{x}{x^2+2x}$
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{1}{x+2}=\frac{1}{2}$
$\lim\limits_{x\to 0^-}\large\frac{|x|}{x^2+2x}$
$\Rightarrow \lim\limits_{x\to 0^-}\large\frac{-x}{x^2+2x}$
$\Rightarrow - \lim\limits_{x\to 0^-}\large\frac{1}{x+2}$
$\therefore \lim\limits_{x\to 0}f(x)\;\text{does not exist}$
Hence (d) is the correct answer.
answered Mar 26, 2014 by balaji
 

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