# 3 part Question -- main problem with limit of arctan

• Oct 20th 2011, 08:02 PM
Robinator
3 part Question -- main problem with limit of arctan
I believe I found the answer to part a), and I also think I could find part c).

I just don't know how to evaluate part b). I was thinking of using the squeeze theorem (or whatever you'd like to call it), but I don't know...

${-\pi \over 2} \leq tan^-1({-16 \over x-2}) \leq {\pi \over 2}$

I don't really know what to do with this
• Oct 20th 2011, 08:08 PM
skeeter
Re: 3 part Question -- main problem with limit of arctan
as $x \to 2^+$ , $\frac{-16}{x-2} \to -\infty$

now check out the graph of $y = \arctan{x}$ ... what value does $y$ approach as $x \to -\infty$ ?
• Oct 21st 2011, 04:45 AM
Robinator
Re: 3 part Question -- main problem with limit of arctan
It is approaching $-\pi \over 2$
• Oct 21st 2011, 06:31 AM
skeeter
Re: 3 part Question -- main problem with limit of arctan
Quote:

Originally Posted by Robinator
It is approaching $-\pi \over 2$

therefore, $\lim_{x \to 2^+} \arctan\left(\frac{-16}{x-2}\right) = \, ?$
• Oct 21st 2011, 07:17 AM
Robinator
Re: 3 part Question -- main problem with limit of arctan
$-\pi \over 2$ Lol. Thank you.