If the function $f\left(x\right) = \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$ is continuous at $x=0$, find $f\left(0\right)$.
$\left[\begin{aligned}
\text{Note: } & \log_e \left(1+x\right) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots \infty \\
& \log_e \left(1-x\right) = -x - \dfrac{x^2}{2} -\dfrac{x^3}{3} - \dfrac{x^4}{4} - \cdots \infty
\end{aligned}\right]$
$\lim\limits_{x \to 0} f\left(x\right)$
$=\lim\limits_{x \to 0} \; \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$
$\lim\limits_{x \to 0} \; \dfrac{\left[\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+ \cdots \infty\right)-\left(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}- \cdots \infty\right)\right]}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x + \dfrac{2x^3}{3}+\dfrac{2x^5}{5}+ \cdots \infty}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)}{x}$
$= 2 \lim\limits_{x \to 0} \; \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)$
$= 2 \times 1 = 2$
Since the function $f\left(x\right)$ is continuous at $x=0$, $\lim\limits_{x \to 0} f\left(x\right)=f\left(0\right)$
$\therefore$ $f\left(0\right)=2$