If mathop {Lim}limits_{x to 0} frac{ln(3 + x) | vedclass

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(A) Given the limit: $\mathop {Lim}\limits_{x 	o 0} \frac{\ln(3 + x) - \ln(3 - x)}{x}$Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$,we get:

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$\frac{2}{3}$

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(A) Given the limit: $\mathop {Lim}\limits_{x 	o 0} \frac{\ln(3 + x) - \ln(3 - x)}{x}$Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$,we get:$\mathop {Lim}\limits_{x 	o 0} \frac{\ln(\frac{3+x}{3-x})}{x}$Using the standard limit $\mathop {Lim}\limits_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,we rewrite the expression:$\mathop {Lim}\limits_{x 	o 0} \frac{\ln(1 + \frac{x}{3}) - \ln(1 - \frac{x}{3})}{x}$$= \mathop {Lim}\limits_{x 	o 0} \left[ \frac{\ln(1 + \frac{x}{3})}{x} - \frac{\ln(1 - \frac{x}{3})}{x} ight]$$= \mathop {Lim}\limits_{x 	o 0} \left[ \frac{1}{3} \cdot \frac{\ln(1 + \frac{x}{3})}{\frac{x}{3}} - (-\frac{1}{3}) \cdot \frac{\ln(1 - \frac{x}{3})}{-\frac{x}{3}} ight]$$= \frac{1}{3}(1) + \frac{1}{3}(1) = \frac{2}{3}$Thus,$k = \frac{2}{3}$.

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

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