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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)

best_answer

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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