# Having trouble taking this seemingly simple limit..

• May 16th 2010, 08:17 AM
charleschafsky
Having trouble taking this seemingly simple limit..
I'm having trouble taking this limit which should definitely be really simple. It looks as follows:

limit as x goes to 1 of (1 - x + lnx)/(1 + cos pi x)

This should probably use l'hopital's rule, since it goes to 0/0 so I take the derivative and get (-1 + 1/x)/(-sin pi x) which equals (-1 + x^-1)/(-sin pi x). This is 0/0 again, so I take l'hopital's rule again. I get (x^-2)/cos pi x. This is -1 as x approaches 1. However, I know this is wrong.

Where have I gone wrong?

Also, how are math equations done on this forum properly? Does it use TeX?
• May 16th 2010, 08:27 AM
alexmahone
Quote:

Originally Posted by charleschafsky
I'm having trouble taking this limit which should definitely be really simple. It looks as follows:

limit as x goes to 1 of (1 - x + lnx)/(1 + cos pi x)

This should probably use l'hopital's rule, since it goes to 0/0 so I take the derivative and get (-1 + 1/x)/(-sin pi x) which equals (-1 + x^-1)/(-sin pi x). This is 0/0 again, so I take l'hopital's rule again. I get (x^-2)/cos pi x. This is -1 as x approaches 1. However, I know this is wrong.

Where have I gone wrong?

Also, how are math equations done on this forum properly? Does it use TeX?

When you differentiate $\displaystyle cos \pi x$ you get $\displaystyle -\pi sin \pi x$. Differentiating $\displaystyle -\pi sin \pi x$ gives $\displaystyle -\pi^2 cos \pi x$.

This forum uses latex for typing math formulas. Quote my post to learn to use it.
• May 16th 2010, 08:42 AM
Soroban
Hello, charleschafsky!

Quote:

$\displaystyle \lim_{x\to1}\frac{1 - x + \ln x}{1 + \cos\pi x}$

This should use L'hopital's rule, since it goes to $\displaystyle \tfrac{0}{0}$.

So I take the derivative and get: .$\displaystyle \frac{-1 + \frac{1}{x}} {-{\color{red}\pi}\sin \pi x}$ . . . . Chain rule!
which equals: .$\displaystyle \frac{-1 + x^{-1}}{-\pi\sin\pi x}$

This is $\displaystyle \tfrac{0}{0}$ again, so I use L'hopital's rule again.

I get: .$\displaystyle \frac{x^{-2}}{{\color{red}\pi^2}\cos\pi x}$

Therefore: .$\displaystyle \lim_{x\to1}\frac{\frac{1}{x^2}}{\pi^2\cos\pi x} \;=\;\frac{\frac{1}{1^2}}{\pi^2\cos\pi} \;=\;\frac{1}{\pi^2(-1)} \;=\;-\frac{1}{\pi^2}$