Evaluating limit

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As it is, the limit does not exist.Of course, if you were to specify the problem precisely as you want it, you might get a different answer. To be precise, you need to specify the codomain of the arctangent function, as it's not injective.Homework Equations: lim(x/y)= lim (x) - lim (y)In summary, the given limit does not exist because the numerator approaches a constant while the denominator approaches 0, which is an undefined or indeterminate form.
  • #1
mathgeek69
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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)
 
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  • #2
mathgeek69 said:

Homework Equations

: lim(x/y)= lim (x) - lim (y)
You cannot mean that!
 
  • #3
mathgeek69 said:
1. lim as x→+pi ##\frac{tan^-1(1/(x-pi))}{pi-x}##
Let [itex]y= x- \pi[/itex] so the problem becomes
[itex]\lim_{y\to 0} \frac{tan^{-1}(1/u)}{u}[/itex]
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)
 
  • #4
haruspex said:
You cannot mean that!

oops, I meant to say
multiplication_law.gif
 
  • #5
mathgeek69 said:
oops, I meant to say [PLAIN]http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/Laws/multiplication_law.gif[/QUOTE] [Broken]
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.
 
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  • #6
HallsofIvy said:
Let [itex]y= x- \pi[/itex] so the problem becomes
[itex]\lim_{y\to 0} \frac{tan^{-1}(1/u)}{u}[/itex]
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 [itex]y= x- \pi[/itex] so the problem becomes
[itex]\lim_{y\to 0} \frac{tan^{-1}(1/-y)}{y}[/itex]

now I do see that the denominator goes to 0 but the numerator does not. This means the limit does not exist ?
 
  • #7
mathgeek69 said:
I don't know where you got the u from. This is what I got:

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

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...
division_law.gif


Because in my answer the denominator reaches 0, the limit of entire question DOES NOT EXIST ?
 
  • #8
mathgeek69 said:

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. :rofl:

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.
 
  • #9
mathgeek69 said:
Let [itex]y= x- \pi[/itex] so the problem becomes
[itex]\lim_{y\to 0} \frac{tan^{-1}(1/-y)}{y}[/itex]

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 [itex]tan^{-1}y^{-1} > \epsilon[/itex].
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.
 

1. What is the purpose of evaluating a limit?

Evaluating a limit is used to determine the behavior of a function as the input approaches a specific value. It helps to identify any potential discontinuities or asymptotes in the graph of the function.

2. How do you evaluate a limit algebraically?

To evaluate a limit algebraically, we first simplify the function as much as possible and then substitute the given value of the input into the expression. If the resulting expression is indeterminate, we can use various algebraic techniques such as factoring or rationalizing the denominator to simplify it further and determine the limit.

3. What is the difference between a one-sided and a two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the given value from one direction (either the left or right). A two-sided limit, on the other hand, looks at the behavior of the function as the input approaches the given value from both directions, and the limit only exists if both one-sided limits are equal.

4. Can a limit exist even if the function is not defined at the given value?

Yes, a limit can exist even if the function is not defined at the given value. This is because a limit only considers the behavior of the function as the input approaches the value, not necessarily at the value itself. However, if the function is not defined at the given value and the limit does not exist, it is considered a discontinuity in the graph of the function.

5. How can limits be used to find the derivative of a function?

Limits are used in the definition of the derivative, which is the instantaneous rate of change of a function at a specific point. By taking the limit of the slope of a secant line as the two points get closer and closer together, we can find the slope of the tangent line at a specific point, which is the derivative of the function at that point.

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