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

1. ## 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...

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

I don't really know what to do with this

2. ## Re: 3 part Question -- main problem with limit of arctan

as $\displaystyle x \to 2^+$ , $\displaystyle \frac{-16}{x-2} \to -\infty$

now check out the graph of $\displaystyle y = \arctan{x}$ ... what value does $\displaystyle y$ approach as $\displaystyle x \to -\infty$ ?

3. ## Re: 3 part Question -- main problem with limit of arctan

It is approaching $\displaystyle -\pi \over 2$

4. ## Re: 3 part Question -- main problem with limit of arctan

Originally Posted by Robinator
It is approaching $\displaystyle -\pi \over 2$
therefore, $\displaystyle \lim_{x \to 2^+} \arctan\left(\frac{-16}{x-2}\right) = \, ?$

5. ## Re: 3 part Question -- main problem with limit of arctan

$\displaystyle -\pi \over 2$ Lol. Thank you.