TITLE: Tedious undefined limit without L'Hospital $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: \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}