Question:
If the function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by
$f(x)= \begin{cases}\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}\right) & \text {, when } x \neq 0 \\ k & , \text { when } x=0\end{cases}$
is continuous, then $\mathrm{k}$ is equal to___________
Solution:
$k=\lim _{x \rightarrow 0}\left(\frac{\ell n(1+3 x)}{x}-\frac{\ell n(1-2 x)}{x}\right)$
$k=3+2=5$