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Homework Help: Why is the limit 2025? a simple plug and chug limit! gone wrong!

  1. Oct 19, 2005 #1
    Hello everyone i have the following prlbem:
    lim (x,y)->(5,-2) (x^5+4x^3y-5xy^2);

    So i let y be fixed at 0 and let x equal 5, and i got 5^5 which is 3125, if i let 0, and y be -2 i get 0. Doesn't this mean the limit doesn't exist? but it says its 2025, ?
  2. jcsd
  3. Oct 19, 2005 #2
    Just plug in 5 for each x value and -2 for each y value, you get 2025. You were doin something completely different.
  4. Oct 19, 2005 #3
    ohhh i did that before and i messed up on multipcation thank u
  5. Oct 20, 2005 #4


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    Homework Helper

    What you're doing in your first post is like saying that if you have:

    [tex]\lim _{x \to 7} x^2[/tex]

    and you plug in x = 3 to get 9 and plug in x = 10 to get 100, then the limit doesn't exist.
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