# How do you find this limit using L`Hopitals rule?

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

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

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Are you sure you were not given the condition $x \to \pi/2^{\color{red}{-}}$? It would be also good for you to show your work. – Pedro Tamaroff May 2 '12 at 22:17
@PeterTamaroff I was given the minus but it looks like it was taken off in the edit ;) – Math_Phase May 2 '12 at 22:17
My fault. I'll fix it. – Pedro Tamaroff May 2 '12 at 22:18
@PeterTamaroff No problem, actually thank you for fixing it for me. It is much more readable now. Now I'm just waiting to see if someone can show me how to do it so I know where I went wrong – Math_Phase May 2 '12 at 22:18
I'm composing an answer right now. – Pedro Tamaroff May 2 '12 at 22:19

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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Thank you for the answer, I'm looking it over and trying to fully understand it. I'll be posting another question (regarding differentials soon) and hopefully I'll have the formatting down by then :) -- Thanks again! – Math_Phase May 2 '12 at 22:39
@Math_Phase You're welcome. – Pedro Tamaroff May 2 '12 at 22:41

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. – Pedro Tamaroff May 2 '12 at 22:37