2
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Write it as $\cos^2 x/(x-\pi/2)$ first. $\endgroup$ – David Mitra Jun 19 '13 at 20:21
  • $\begingroup$ Try substituting $u = x-\pi/2$, noting that $\sec(x)=-\csc(u)$ $\endgroup$ – Omnomnomnom Jun 19 '13 at 20:24
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.