If the function f defined on (- (1)/(3), 1 / 3) by f(x) = \ cases...

\lim_{x \to 0} f\left(x\right) = \lim_{x \to 0} \left(\dfrac{\ln \left(1 + 3x\right)}{x} - \dfrac{\ln \left(1 - 2x\right)}{x}\right)$<br />$= \lim_{x \to 0} \left(\dfrac{3\ln \left(1 + 3x\right)}{3x} + \dfrac{2\ln \left(1 - 2x\right)}{- 2x}\right) = 3 + 2 = 5$</p><p>\left[\lim_{f\left(x\right) \to 0}\dfrac{\ln \left\{1 + f\left(x\right)\right\}}{f\left(x\right)} = 1\right]$

$∴ f\left(x\right)$, will be continuous if $f\left(0\right) = \lim_{x \to 0} f\left(x\right)$

Thus, $k = 5$