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$\lim_{x \to \pi/2^{-}} { \tan(x) \over \ln \left(\frac{\pi}{2} - x\right)}$

I attempted to do it but I keep getting $0/-1$

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    For future considerations, it will be good that you show what you have done to solve the problem, since maybe a simple tweak in your work might solve the problem. You might also want to learn some LaTeX usage [here](http://www.andy-roberts.net/writing/latex/mathematics_2) and [here](http://www.andy-roberts.net/writing/latex/mathematics_1).2012-05-02

2 Answers 2

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You need to evaluate

$\lim_{x \to \pi/2 ^-} { \tan(x) \over \log \left(\frac{\pi}{2} - x\right)}$

Not this is the same as evaluating $y=\pi/2-x$ as $y \to 0^+$, viz:

$\lim_{y \to 0^+} { \tan \left( \frac {\pi} 2-y\right) \over \log y}$

Note that $\tan \left( \frac {\pi} 2-y\right)=\dfrac 1 {\tan y}$, so the limit you are looking for is:

$\lim_{y \to 0^+}{\cot y \over {\log y}}$

This is an indeterminate $\infty \over \infty$ form, so we'll aplly L'Hòpital's rule:

$\lim_{y \to 0^+}{\cot y \over {\log y}} = -\lim_{y \to 0^+}{\sin^{-2} y \over { y^{-1}}} $

which is the same as

$-\lim_{y \to 0^+}{y \over {\sin y}}{ 1 \over \sin y} $ $ - \mathop {\lim }\limits_{y \to {0^ + }} \underbrace {\frac{y}{{\sin y}}}_{ \to 1}\frac{1}{{\underbrace {\sin y}_{ \to 0}}} = - \infty $

As André is suggesting in his answer, the function does not approach a finite limit, so formally we say the limit does not exist. However, as you might have experienced, we might informally note that the function takes larger and larger negative values for $x$ near $\pi /2$ by

$\lim_{x \to \pi/2 ^-} { \tan(x) \over \log \left(\frac{\pi}{2} - x\right)}=-\infty$

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    @Math_Phase You're welcome.2012-05-02
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Apply L'Hospital's Rule. So we want the limit of $\frac{\sec^2 x}{-\dfrac{1}{\frac{\pi}{2}-x}}$ as $x$ approaches $\pi/2$ from the left. Rewrite the above expression as $-\frac{\frac{\pi}{2}-x}{\cos^2 x}.$ Apply L'Hospital's Rule again. Taking derivatives we arrive at $\frac{1}{-2\sin x\cos x}.$ The limit of this, as $x$ approaches $\pi/2$ from the left, does not exist, and therefore neither does our original limit.

Or else, if we allow $\infty$ and $-\infty$ as limits, then as $x$ approaches $\pi/2$ from the left, $\frac{1}{-2\sin x\cos x}$ approaches $-\infty$, since $\sin x$ approaches $1$, and $\cos x$ approaches $0$ through positive values. So the limit of the original expression is $-\infty$.

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    Andre, I side by you that $\infty$ is not really a limit, but maybe it would be good to clarify that issue here.2012-05-02