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The function $f:R\{0\}\rightarrow R$ given by $f(x)=\large\frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at $x=0$ by defining f(0) as

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;-1$

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Given :$f(x)=\large\frac{1}{x}-\frac{2}{e^{2x}-1}$
$\Rightarrow f(0)=\lim\limits_{x\to 0}\large\frac{1}{x}-\frac{2}{e^{2x}-1}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{(e^{2x}-1)-2x}{x(e^{2x}-1)}\qquad\bigg[\large\frac{0}{0}\bigg]$ form
$f(0)=\lim\limits_{x\to 0}\large\frac{4e^{2x}}{2(xe^{2x}2+e^{2x}.1)+e^{2x}.2}$
$\Rightarrow\lim\limits_{x\to 0}\large\frac{4e^{2x}}{4xe^{2x}+2e^{2x}+2e^{2x}}$
$\Rightarrow\lim\limits_{x\to 0}\large\frac{4e^{2x}}{4(xe^{2x}+e^{2x})}$
$\Rightarrow\lim\limits_{x\to 0}\large\frac{4e^{\large 0}}{4(0+e^{\large 0})}$
$\Rightarrow 1$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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