If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by
$f(x)=\left\{\begin{array}{l}\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}\right), \text { when } x \neq 0 \\ k, \text { when } x=0\end{array}\right.$ is continuous, then $k$ is equal to_____.
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{1}{x} \ln \left(\frac{1+3 x}{1-2 x}\right)\right)$
$=\lim _{x \rightarrow 0}\left(\frac{\ln (1+3 x)}{x}-\frac{\ln (1-2 x)}{x}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{3 \ln (1+3 x)}{3 x}-\frac{2 \ln (1-2 x)}{-2 x}\right)$
$=3+2=5$
$\because f(x)$ will be continuous
$\therefore \quad \mathrm{k}=f(0)=\lim _{x \rightarrow 0} f(x)=5$
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