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The function $f:R/\left \{ 0 \right \}\rightarrow R$ given by

$f\left ( x \right )= \frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at $x= 0$ by defining $f \left ( 0 \right )$ as

Option 1)
$0$

Option 2)
$1$

Option 3)
$2$

Option 4)
$-1$

Answers (1)

best_answer

As we learnt in

Evaluation of Exponential Limits -

$\lim_{x\rightarrow 0}\:\:\:\frac{e^{x}-1}{x}=1$

$\lim_{x\rightarrow 0}\:\:\:\frac{a^{x}-1}{x}=log_{e}{a}$

$\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{x}\neq 1$

$\therefore \:\:\:\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{x}\times \frac{K}{K}$

$\therefore \:\:\:K\lim_{x\rightarrow 0}\:\:\:\frac{e^{Kx}-1}{Kx}=K\times 1=K$

- wherein

$\lim_{x\rightarrow 0}\:\:\:\frac{e^{x}-1}{x}$

$x\:must\:be\:same$

$\lim_{x\rightarrow 0}\frac{1}{x}-\frac{2}{e^{2x}-1}$ $= \lim_{x\rightarrow 0}\frac{e^{2x}-1-2x}{(e^{2x}-1)}$

We know that

$e^{x} = 1+\frac{x}{1!}+ \frac{x^2}{2!}+\frac{x^3}{3!}+.................$

Put $x\rightarrow 2x$

$e^{2x} = 1+\frac{2x}{1!}+ \frac{4x^2}{2!}+\frac{8^3}{3!}+.................$

$\therefore e^{2x}-1-2x = 2x^2+{\frac{8x^3}{3!}}+...................$

$\therefore \frac{e^{2x}-1-2x}{x(e^{2x}-1)} = \frac{2x+{\frac{8x^3}{3!}}+...................}{2x+2x^2+{\frac{8x^3}{3!}}+...................}$ $=\frac{2}{2}=1$

Option 1)

$0$

Incorrect

Option 2)

$1$

Correct

Option 3)

$2$

Incorrect

Option 4)

$-1$

Incorrect

Posted by

prateek

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