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)

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

Exams
Articles
Questions