How Does the Limit of tan^-1(1/(x-pi))/(pi-x) Behave as x Approaches pi?

[mathgeek69]

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

Homework Equations

: lim(x/y)= lim (x) - lim (y)

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)

 

[haruspex]

Homework Equations

: lim(x/y)= lim (x) - lim (y)

You cannot mean that!

 

[HallsofIvy]

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

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.

Homework Equations

: lim(x/y)= lim (x) - lim (y)

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

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)

 

[mathgeek69]

You cannot mean that!

oops, I meant to say

 

[haruspex]

oops, I meant to say [PLAIN]http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/Laws/multiplication_law.gif[/QUOTE]
OK, but doesn't help if the two limits on the RHS are contradictory, e.g. one goes to infinity while the other goes to zero. However, that's not the case here. Follow HallsofIvy's hint.

 

[mathgeek69]

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 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 ?

 

[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 ?

The limit division law says if lim of f(x) and g(x) exists and if lim of g(x) isn't 0 then...

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

 

[Mandelbroth]

Homework Equations

: lim(x/y)= lim (x) - lim (y)

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.

Joking aside, to answer your question of your limit's existence, it should not exist.

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

 

[haruspex]

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 ?

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:

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

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.