If the function f defined on  $\left(-\frac{1}{3},\frac{1}{3}\right)$by $f\left(x\right)=\left\{\begin{array}{l}\frac{1}{x}{\log }_e\left(\frac{1+3x}{1-24}\right),whenx\ne 0\\ k,whenx=0\end{array}\right.$ is continuous, then k is equal to _____

$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{1}{x}\ln \left(\frac{1+3x}{1-2x}\right)\right)=\underset{x\to 0}{\lim }\left(\frac{\ln \left(1+3x\right)}{x}-\frac{\ln \left(1-2x\right)}{x}\right)$

                                                 $$\begin{gathered} =\underset{x\to 0}{\lim }\left(\frac{3\ln \left(1+3x\right)}{3x}-\frac{2\ln \left(1-2x\right)}{-2x}\right) \\ =3+2 \\ =5 \end{gathered}$$

Since f(x) is  continuous at x=0, we have $f\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)$

$\Rightarrow k=5$