If mathop {lim }limits_{x to 0} frac{{log (3 | vedclass

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(C) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k$. Applying $L'Hospital's$ rule,we differentiate the

Vedclass · Accepted Answer

$\frac{2}{3}$

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(C) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k$. Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$: $\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(\log(3+x) - \log(3-x))}{\frac{d}{dx}(x)} = k$. $\mathop {\lim }\limits_{x 	o 0} \frac{\frac{1}{3+x} - (\frac{1}{3-x} 	imes -1)}{1} = k$. $\mathop {\lim }\limits_{x 	o 0} (\frac{1}{3+x} + \frac{1}{3-x}) = k$. Substituting $x = 0$: $\frac{1}{3+0} + \frac{1}{3-0} = k$. $\frac{1}{3} + \frac{1}{3} = k$. $k = \frac{2}{3}$.

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k,$ then the value of $k$ is

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