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 $x=0$?
 A: Hint: for every real number $c$, show that $g_c(x)=\log(1+cx)/x$ has a limit $\ell(c)$ when $x\to0$ and compute $\ell(c)$. Then apply this partial result to your question.

Here is an alternative solution. First,
$$
f(x)=\int_{\,-b}^{\,\,a}\,\frac{\mathrm{d}z}{1+zx}.
$$
Second, for small enough values of $|x|$, $|1/(1+zx)|\le1/(1-c)$ with $c=|x|\max\{|a|,|b|\}$ and $c<1$. Hence dominated convergence applies when $x\to0$, which means that $f(x)$ converges to the integral of the limit of the function. The latter is $1$ for every $z$ hence its integral is $a+b$.
A: 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.
A: As Chandru has pointed out, this is the $\ln(\cos\,x)/x$ problem all over again. So, being unimaginative at this time of the night, I will solve it in the same way all over again.  Please note that most of the time in this post, $f$ does not refer to your $f$. So let's call your function $g$. The value we need to assign to it at $0$ in order to end up with a continuous function is
$$\lim_{x\to 0}\frac{g(x)}{x}$$
if this limit exists. (If the limit does not exist, no assignment of value to $g$ at $0$ will make the resulting function continuous.)
Using the notation of Did's answer, let's find
$$\lim_{x \to 0} \frac{\ln(1+cx) -\ln(1)}{x-0}.$$
We have done nothing here, since $\ln 1=0$, and $x-0=x$, but we have done nothing in what will turn out to be a useful way.
We recognize the above expression as the definition of $f'(0)$, where $f(x)=\ln(1+cx)$.
For our particular function $f$, we have
$$f'(x)=\frac{c}{1+cx}$$ 
and therefore $f'(0)=c$.
Thus the required answer is $a-(-b)$, which is $a+b$.
Comment: The limit is also easily arrived at by considering the power series expansion of $\ln(1+u)$.
We sort of took a chance in the above calculation (well, not really, sine the answer is obvious from the power series).
In principle, it would have been better to do exactly the same thing, but with $f(x)=\ln(1+ax) -\ln(1-bx)$.
We would then have 
$$f'(x)=\frac{a}{1+ax}-\frac{-b}{1-bx}$$
and again we would obtain $f'(0)=a+b$.
The method is available for any limit of the kind 
$$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$
once we know how to differentiate $f$.
It can be thought of as a hypersimple version of L'Hospital's Rule.  Or else, more ambitiously but less accurately, as the beginnings of an explanation of why the L'Hospital's Rule works.  
