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Question

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


A
2
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B
1
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C
0
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D
1
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Solution

The correct option is D $$1$$
Given$$\displaystyle f\left( x \right) =\frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } $$
$$ \displaystyle \Rightarrow f\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \left\{ \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 }  \right\}  } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 2x }-1-2x }{ x\left( { e }^{ 2x }-1 \right)  }  }  \ ....... \quad \left[ \frac { 0 }{ 0 } form \right] $$
$$\therefore $$ usingL'Hospital rule
$$f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad  }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) }  } $$
$$\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } }  } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right)  } =1$$

Mathematics

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