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 $0/-1$
 A: 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 
$$\lim_{x \to \pi/2 ^-}  { \tan(x) \over  \log \left(\frac{\pi}{2} - x\right)}=-\infty$$
A: 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 $-\infty$. 
