# Evaluating limit

1. May 29, 2013

### mathgeek69

1. lim as x→+pi $\frac{tan^-1(1/(x-pi))}{pi-x}$

2. Relevant equations: lim(x/y)= lim (x) - lim (y)

3. The attempt at a solution:

umm don't know where to go from here...

lim as x→+pi [tan^-1(1/(x-pi))] - lim as x→+pi (pi-x)

Last edited: May 29, 2013
2. May 30, 2013

### haruspex

You cannot mean that!

3. May 30, 2013

### HallsofIvy

Staff Emeritus
Let $y= x- \pi$ so the problem becomes
$\lim_{y\to 0} \frac{tan^{-1}(1/u)}{u}$
The denominator is going to 0 while the numerator is not.

I assume this is a misprint. But even "lim(x/y)= lim(x)/lim(y)" is not generally true.

4. May 30, 2013

### mathgeek69

oops, I meant to say

5. May 30, 2013

### haruspex

Last edited by a moderator: May 6, 2017
6. May 30, 2013

### mathgeek69

I don't know where you got the u from. This is what I got:

Let $y= x- \pi$ so the problem becomes
$\lim_{y\to 0} \frac{tan^{-1}(1/-y)}{y}$

now I do see that the denominator goes to 0 but the numerator does not. This means the limit does not exist ?

7. May 30, 2013

### mathgeek69

The limit division law says if lim of f(x) and g(x) exists and if lim of g(x) isnt 0 then...

Because in my answer the denominator reaches 0, the limit of entire question DOES NOT EXIST ?

8. May 30, 2013

### Mandelbroth

In a less formal setting, I'd be laughing hysterically with iterated references to a time paradox internet meme. In this case, though, dying of laughter on the inside will have to suffice. :rofl:

The numerator should approach $-\pi/2$ while your denominator approaches 0, which generally implies the limit is considered to be undefined.

9. May 30, 2013

### haruspex

Yes. Given ε > 0, you can find arbitrarily small y such that $tan^{-1}y^{-1} > \epsilon$.
However, this in your later post is not quite correct:
You also need the fact that the numerator does not converge to 0. If both converged to zero you would need a subtler approach to resolve the matter.