# Limit: $\lim\limits_{x\to{\pi/2}^+} \frac{\ln(x-\pi/2)}{\tan(x)}$ using De l'Hôpital's rule?

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.

• Write it as $\cos^2 x/(x-\pi/2)$ first. – David Mitra Jun 19 '13 at 20:21
• Try substituting $u = x-\pi/2$, noting that $\sec(x)=-\csc(u)$ – Omnomnomnom Jun 19 '13 at 20:24

$$\lim_{x \to \pi/2+} \frac{{1/(x-\pi/2)}}{\sec^2(x)}=\lim_{x \to \pi/2+} \frac{\cos^2(x) }{ x-\pi/2 }=\lim_{u \to 0+} \frac{\cos^2(u+\pi/2) }{ u }$$$$=\lim_{u \to 0+} \frac{\sin^2(u ) }{ u }=\lim_{u \to 0+} \frac{2 \sin(u )\cos(u) }{ 1 }=0.$$