If \lim_{x\rightarrow 0}\frac{\log \left ( 3+x \right )-\log \left ( 3-x \right )}{x}= k     the value of k is

  • Option 1)

    -1/3

  • Option 2)

    2/3

  • Option 3)

    -2/3

  • Option 4)

    0

 

Answers (1)

As we learnt in 

Evaluation of Logarithmic Limits -

\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+x)}{x}=1

\Rightarrow \lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{x}\neq 1

but\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{x}\times \frac{K}{K}

\therefore \:\:\:K\lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1+Kx)}{Kx}

=K\times 1=K

\Rightarrow \lim_{x\rightarrow 0}\:\:\:\frac{log_{e}(1-x)}{x}=-1

- wherein

\frac{1+x}{x}

x\:must\:be\:same

 

 \lim_{n \to 0} \: \: \frac{log(3+x)-log(3-x)}{x}=k

\lim_{n \to 0} \: \: \frac{log3(1+\frac{x}{3})-log3(1-\frac{x}{3})}{x}

\lim_{n \to 0} \: \: \frac{log3+log(1+\frac{x}{3})-log3-log(1-\frac{x}{3})}{x}

\lim_{n \to 0} \: \: \frac{log(1+\frac{x}{3})-log(1-\frac{x}{3})}{x}

\Rightarrow \lim_{n \to 0} \: \: \frac{log(1+\frac{x}{3})}{3\times \frac{x}{3}} \: + \lim_{n \to 0} \: \: \frac{log(1-\frac{x}{3})}{-3\times \frac{x}{3}}

=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}

 


Option 1)

-1/3

This option is incorrect

Option 2)

2/3

This option is correct

Option 3)

-2/3

This option is incorrect

Option 4)

0

This option is incorrect

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