Find the limit
$$\lim_{x \to {\pi/2}^+} \frac{\ln(x-\pi/2)}{\tan(x)}$$
So, both the numerator and the denominator approach negative infinity. There De l'Hôpital's rule applies. I found the derivative of both and got:
$$\lim_{x \to {\pi/2}^+} \frac{{1/(x-\pi/2)}}{\sec^2(x)}$$
I keep on applying De l'Hôpital's rule yet still cannot find an expression that will not result in zero as the denominator.
Best Answer
$$\lim_{x \to \pi/2+} \frac{{1/(x-\pi/2)}}{\sec^2(x)}=\lim_{x \to \pi/2+} \frac{\cos^2(x) }{ x-\pi/2 }=\lim_{u \to 0+} \frac{\cos^2(u+\pi/2) }{ u }$$$$=\lim_{u \to 0+} \frac{\sin^2(u ) }{ u }=\lim_{u \to 0+} \frac{2 \sin(u )\cos(u) }{ 1 }=0.$$