calculus - 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?

Answer

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.