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

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(A) આપેલ લક્ષ: $\mathop {Lim}\limits_{x 	o 0} \frac{\ln(3 + x) - \ln(3 - x)}{x}$$\ln(a) - \ln(b) = \ln(\frac{a}{b})$ ગુણધર્મનો ઉપયોગ કરતા:$\mathop {L

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) આપેલ લક્ષ: $\mathop {Lim}\limits_{x 	o 0} \frac{\ln(3 + x) - \ln(3 - x)}{x}$$\ln(a) - \ln(b) = \ln(\frac{a}{b})$ ગુણધર્મનો ઉપયોગ કરતા:$\mathop {Lim}\limits_{x 	o 0} \frac{\ln(\frac{3+x}{3-x})}{x}$પ્રમાણિત લક્ષ $\mathop {Lim}\limits_{u 	o 0} \frac{\ln(1+u)}{u} = 1$ નો ઉપયોગ કરતા:$\mathop {Lim}\limits_{x 	o 0} \frac{\ln(1 + \frac{x}{3}) - \ln(1 - \frac{x}{3})}{x}$$= \mathop {Lim}\limits_{x 	o 0} \left[ \frac{\ln(1 + \frac{x}{3})}{x} - \frac{\ln(1 - \frac{x}{3})}{x} ight]$$= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$આમ,$k = \frac{2}{3}$.

જો $\mathop {Lim}\limits_{x \to 0} \frac{\ln(3 + x) - \ln(3 - x)}{x} = k$ હોય,તો $k$ ની કિંમત શોધો.

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