Tedious undefined limit without L'Hospital $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$ When I try to calculate this limit:
$$\mathop {\lim }\limits_{x \to \frac{\pi}{2}^+} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$$
I find this:
$$\begin{array}{l}
L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}^+} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}\\
\text{variable changing}\\
y = 2x - \pi \\
x \to \frac{\pi }{2}\,\,\,\, \Rightarrow \,\,\,y \to 0\\
\text{so:}\\
L = \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\tan \,\left( {\frac{{y + \pi }}{2}} \right)}}{{\ln \,(y)}} = \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\tan \,\left( {\frac{y}{2} + \frac{\pi }{2}} \right)}}{{\ln \,(y)}}\\
 = \mathop {\lim }\limits_{y \to 0} \,\,\frac{{ - \cot\,\left( {\frac{y}{2}} \right)}}{{\ln \,(y)}} =  - \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\csc (y) + \cot (y)}}{{\ln \,(y)}}\\
 = \frac{{ \pm \infty  \pm \infty }}{{ - \infty }} = ??
\end{array}$$
and in the latter part I get stuck,
should be obtained using mathematical software $L= \pm \infty$ 
how I justify without L'Hospital?
 A: 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.
A: \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}
