The function [f:R/ 0 R] given by [f( x ) = 1 x - 2 e^2x - 1] can be made continuous at x =…

Question

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

Accepted Answer

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 $$\left. {\,\,} \right]$$

$$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4\left( {x{e^{2x}} + {e^{2x}}} \right)}} = {{4.{e^0}} \over {4\left( {0 + {e^0}} \right)}} = 1$$

Solutions · 1