Tedious undefined limit without L'Hospital $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$

The change of variables is a good start! Write $$ -\frac{\cot \frac y2}{\ln y} = -2 \cos\frac y2 \cdot \frac{\frac y2}{\sin\frac y2} \cdot \frac1{y\ln y}. $$ The first factor has limit $-2$ as $y\to0$, by continuity; the second factor has limit $1$ as $y\to0$, due to the fundamental limit result $\lim_{x\to0} \frac{\sin x}x = 1$; and the denominator of the last factor tends to $0$ as $y\to0+$ (and is undefined as $y\to0-$). Therefore the whole thing tends to $-\infty$.

This depends upon two fundamental limits, namely $\lim_{x\to0} \frac{\sin x}x = 1$ and $\lim_{x\to0+} x\ln x = 0$. The first can be established by geometrical arguments, for sure. I'd have to think about the second one, but presumably it has a l'Hopital-free proof as well.

\begin{align} \color{#66f}{\large\lim_{x\ \to\ \pi/2}{\tan\left(x\right) \over \ln\left(2x - \pi\right)}}&= \lim_{y\ \to\ 0}{\tan\left(y/2 + \pi/2\right) \over \ln\left(y\right)} =\lim_{y\ \to\ 0}{-\cot\left(y/2\right) \over \ln\left(y\right)} =-2\lim_{y\ \to\ 0}{1 \over y\ln\left(y\right)}\,{y/2 \over \tan\left(y/2\right)} \\[5mm]&=-2\left[\lim_{y\ \to\ 0}{1 \over y\ln\left(y\right)}\right]\ \underbrace{\left[\lim_{y\ \to\ 0}{y/2 \over \tan\left(y/2\right)}\right]} _{\displaystyle=\color{#c00000}{\large 1}} =-2\lim_{y\ \to\ 0}{1 \over \ln\left(y^{y}\right)} = \color{#66f}{\large +\infty} \end{align}