5
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Interesting problem. Why do you think there is a solution without L'Hospital? $\endgroup$ – Cheerful Parsnip Oct 21 '14 at 22:40
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ For $x\ln x$, write $x = e^{-t}$ and use that $e^t > t^2/2$ for $t \geqslant 0$. $\endgroup$ – Daniel Fischer Oct 21 '14 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? $\endgroup$ – Greg Martin Oct 22 '14 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. $\endgroup$ – Daniel Fischer Oct 22 '14 at 17:41
  • $\begingroup$ Cool, you've convinced me! $\endgroup$ – Greg Martin Oct 22 '14 at 18:03
1
$\begingroup$

\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}

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.