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Homework Help: Multi variable Limit

  1. Jul 12, 2018 #1
    1. The problem statement, all variables and given/known data
    [​IMG] https://gyazo.com/268bef206850bfbf30fb0cca3f783599 <----- The question
    2. Relevant equations


    3. The attempt at a solution
    Had this on a test today, honestly not sure how to evaluate. I know you can pass the limit to the inside of arctan but I can't see how the inside goes to infinity. Any help would be appreciated I just want to understand how to do it!
     
  2. jcsd
  3. Jul 12, 2018 #2

    scottdave

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    I don't get infinity for the inside, either. What is the answer that they gave? Did they get pi/2 for the limit? What answer did you get for the inside (and the limit)?
     
  4. Jul 12, 2018 #3
    the answer given is pi/2 for the overall answer and +infinity for the inside
     
  5. Jul 12, 2018 #4
    The only advice I have received so far is to take note that the bottom is the square of the distance from (0,1) but I can’t figure out what they mean
     
  6. Jul 12, 2018 #5

    Mark44

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    The numerator inside the brackets is approaching 1, and the denominator is approaching 0 from the positive side. It seems pretty clear to me that the part in brackets is approaching ##+\infty##, so the overall limit is what?
     
  7. Jul 13, 2018 #6
    the way I had learned it was that the only way to prove a limit existed was through methods such as delta-epsilon definition, switching to R or Rho, or by squeeze theorem. Is there any mathematical way to show that this goes to +infinity? We were told we were unable to evaluate limits by observation.
     
  8. Jul 13, 2018 #7

    Mark44

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    For the record, here is the limit in the image link you posted:
    $$\text{Find }\lim_{(x, y) \to (0, 1)} \tan^{-1}\left[\frac{x^2 + 1}{x^2 + (y - 1)^2}\right]$$
    The problem asks you to find the limit, but doesn't ask you to prove that this value is the limit. Proving the limit would require the use of a delta-epsilon proof or a theorem such as the squeeze theorem.

    A major difficulty with these kinds of limits is when the limit is an indeterminate form such as ##[\frac 0 0]##. This limit is not one of the indeterminate forms, since the numerator approaches 1 and the denominator approaches 0 through the positive numbers.
     
  9. Jul 13, 2018 #8

    Ray Vickson

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    Yes, the way to do it is to use the definition of the notion ##(x,y) \to (0,1)##. Just using that definition (and a few simple manipulations), it is easy to show that for any (large) number ##N > 0##, we know that as ##(x,y)## nears ##(0,1)## (in a well-defined sense that you are supposed to know) we have that the denominator of the inside fraction becomes ##< 1/N##. The numerator is ## \geq 1##, so the inside fraction is ##> N##. That is essentially the definition of ##\text{fraction} \to +\infty.##
     
  10. Jul 13, 2018 #9

    scottdave

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    Oh, that's easier to read than the tiny image. I misread what x and y we're approaching.
     
  11. Jul 13, 2018 #10

    epenguin

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    It doesn't matter whether the 'inside' is +∞ or -∞ , its tan-1 is π/2 in either case.

    You appear to be able to approach the limit in several ways including from y>1.

    The denominator is indeed the squared distance of a point (x, y) from (0,1).
     
  12. Jul 14, 2018 #11

    Mark44

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    Not so. ##\lim_{x \to -\infty}\tan^{-1}(x) = -\frac \pi 2##. For this inverse trig function, the restricted-domain function ##y = \tan(x), -\frac \pi 2 < x < \frac \pi 2## is used.
     
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