Continuity of this function at $x=0$ The following function is not defined at $x=0$:
$$f(x) = \frac{\log(1+ax) - \log(1-bx)}{x} .$$
What would be the value of $f(0)$ so that it is continuous at $x=0$?

Do you want to evaluate the limit at $0$. Then you can see that 
\begin{align*}
f(x) = \lim_{x \to 0} \biggl(\frac{1+ax}{1-bx}\biggr)^{1/x} &=\lim_{x \to 0} \biggl(1+ \frac{(b+a)x}{1-bx}\biggr)^{1/x} \\\ &=\lim_{x \to 0} \biggl(1+ \frac{(b+a)x}{1+bx}\biggr)^{\small \frac{1+bx}{(b+a)x} \cdot \frac{(b+a)x}{x \cdot (1+bx)}} \\\ &=e^{\small\displaystyle\tiny\lim_{x \to 0} \frac{(b+a)x}{x\cdot (1+bx)}} = e^{b+a} \qquad \qquad \Bigl[ \because \small \lim_{x \to 0} (1+x)^{1/x} =e \Bigr]
\end{align*}
Therefore as $x \to 0$, $\log(f(x)) \to (b+a)$
Please see this post: Solving $\lim\limits_{x \to 0^+} \frac{\ln[\cos(x)]}{x}$ as a similar kind of methodology is used to solve this problem.
