Another simple limit problem L'Hôpital tells us that the following limit is $\frac{\pi}{2}$, but I'm wondering if there is a smart way of evaluating this limit without using L'Hôpital
$$ \lim_{x \to \frac{1}{4}}\frac{\log(\tan(\pi x))}{4x - 1} =?$$

Completing the solution of the lab, we have using infinites series
$$
\frac{1}{1 + h} = 1 - h + h^2 - \ldots \quad |h|<1 \quad \Rightarrow
$$
$$
\int\frac{1}{1 + h}dh = \ln(1 + h) + C = h - \frac{h^2}{2} + \frac{h^3}{3} - \ldots
$$
For $h = 0 \ \Rightarrow \ C = 0$. Futhermore, $\tan x = x + x^3/3 + \ldots$. Thus, 
$$
\lim_{y \to 0}\frac{\ln\biggl(1 + \tan \frac{\pi y}{4}\biggr)}{y} = \lim_{y \to 0}\frac{\tan\frac{\pi y}{4} - \frac{1}{2}\tan^2 \frac{\pi y}{4} + \ldots}{y} =  \lim_{y \to 0} \frac{\frac{\pi y}{4} + O(y^2)\ldots}{y} = \frac{\pi}{4}
$$
Similarly,
$$
\lim_{y \to 0}\frac{\ln\biggl(1 - \tan \frac{\pi y}{4}\biggr)}{y} = -\frac{\pi}{4}
$$
Thus,
$$
\lim_{x \to 1/4}\frac{\ln(\tan(\pi x))}{4x - 1} = \frac{\pi}{2}
$$
