Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I understand how to use L'Hopitals rule for the most part but these two problems confuse me to no end. I would appreciate it if someone could show me how they are to be done.

first one...


second one...

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

share|cite|improve this question
Write the first one as ${x\over e^x}$; can you do it now? – Gerry Myerson Nov 29 '12 at 3:22
Part of your difficulty with the second one may be that you’ve miscopied it: it must be $$\lim_{x\to\frac{\pi}2^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}$$ rather than what you have. – Brian M. Scott Nov 29 '12 at 3:25
@GerryMyerson That would bring it back to 1/e^x which is 0? – DoesTheLimExist Nov 29 '12 at 3:32
@Brian On my homework it is written how I placed it up above. (I got it wrong -- I couldn't narrow it down to a solution) – DoesTheLimExist Nov 29 '12 at 3:33
Then there’s an error in the homework, because when $x$ is a little less than $\pi/2$, $\frac{x}2-x$ is negative, and its natural log isn’t even defined. – Brian M. Scott Nov 29 '12 at 3:35
up vote 1 down vote accepted

Gerry Myerson’s comment should take care of the first problem. The second one has to be misstated: I’ve very little doubt that it should be

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

Corrected: If so, apply l’Hospital’s rule once and do a little simplification:

$$\begin{align*} \lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}&=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\sec^2 x}{\frac{-1}{\frac{\pi}2-x}}\\\\ &=\lim_{x\to\left(\frac{\pi}2\right)^-}\sec^2 x\left(x-\frac{\pi}2\right)\\\\ &=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{x-\frac{\pi}2}{\cos^2 x}\;. \end{align*}$$

Now apply l’Hospital’s rule one more time.

share|cite|improve this answer
The last step in derivation should be $\lim_{x\to \frac{\pi}{2}^{-}} (x-\frac{\pi}{2}){\sec(x)}^2 $. – Mhenni Benghorbal Nov 29 '12 at 3:47
@Brian That would be the numerator 1/0?? Making it undefined. How did you get rid of the -1 in the last part of that? – DoesTheLimExist Nov 29 '12 at 3:47
@MhenniBenghorbal Wouldn't that make the limit be approaching 0? – DoesTheLimExist Nov 29 '12 at 3:49
@DoesTheLimExist: The limit goes to $-\infty$. – Mhenni Benghorbal Nov 29 '12 at 3:52
@Mhenni: Thanks; fixed. – Brian M. Scott Nov 29 '12 at 3:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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