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?