JEE MAIN - Mathematics (2007 - No. 2)
The function $$f:R/\left\{ 0 \right\} \to R$$ given by
$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
can be made continuous at $$x$$ = 0 by defining $$f$$(0) as
$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
can be made continuous at $$x$$ = 0 by defining $$f$$(0) as
0
1
2
$$-1$$
Explanation
Given, $$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
$$ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$$ $$\left[ \, \right.$$ $${0 \over 0}$$ form $$\left. \, \right]$$
$$\therefore$$ using, $$L$$'Hospital rule
$$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {2\left( {x{e^{2x}}2 + {e^{2x}}.1} \right) + {e^{2x}}.2}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,$$ $$\left[ {\,\,} \right.$$ $${0 \over 0}$$ form $$\left. {\,\,} \right]$$
$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4\left( {x{e^{2x}} + {e^{2x}}} \right)}} = {{4.{e^0}} \over {4\left( {0 + {e^0}} \right)}} = 1$$
$$ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$$ $$\left[ \, \right.$$ $${0 \over 0}$$ form $$\left. \, \right]$$
$$\therefore$$ using, $$L$$'Hospital rule
$$f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {2\left( {x{e^{2x}}2 + {e^{2x}}.1} \right) + {e^{2x}}.2}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,$$ $$\left[ {\,\,} \right.$$ $${0 \over 0}$$ form $$\left. {\,\,} \right]$$
$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4\left( {x{e^{2x}} + {e^{2x}}} \right)}} = {{4.{e^0}} \over {4\left( {0 + {e^0}} \right)}} = 1$$
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