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?

  • $\begingroup$ Interesting problem. Why do you think there is a solution without L'Hospital? $\endgroup$ Oct 21, 2014 at 22:40

2 Answers 2


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.

  • $\begingroup$ For $x\ln x$, write $x = e^{-t}$ and use that $e^t > t^2/2$ for $t \geqslant 0$. $\endgroup$ Oct 21, 2014 at 23:03
  • $\begingroup$ True, but that inequality's proof is really l'Hopital in disguise in some sense. Is there a derivative-free proof? $\endgroup$ Oct 22, 2014 at 16:11
  • $\begingroup$ I'd disagree. $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ is for me the definition of the exponential function [much more direct and elementary than "the solution of $y' = y$ with $y(0) = 1$"], and the inequality is immediate from that definition. If you start from a different definition of the exponential function, there may be other easy proofs. $\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$ gives you the lower bound $x^2/2$ (for $x \geqslant 0$) also directly from binomial expansion. For the differential equation definition, however, I don't see a derivative-free proof. $\endgroup$ Oct 22, 2014 at 17:41
  • $\begingroup$ Cool, you've convinced me! $\endgroup$ Oct 22, 2014 at 18:03

\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}


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