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Multivariable Limits assignment

  1. May 22, 2008 #1
    I can't get this problem right and it's part of a web-based assignment that I have to submit. In order to get credit for the problem, all of my answers need to be right; Ive tried many different times and I cant seem to figure out what I am doing wrong. Here it is with my explanations as to why I think I'm right; please, can someone find my mistake?

    Find the following limits. If a limit diverges to infinity, state your answer as "INF" (without the quotation marks). If it diverges to negative infinity, state your answer as "MINF".If it diverges without being infinity or negative infinity, state your answer as "DIV".

    lim(x,y)->(0,0) (3x^2 + y^2)/(x^4 + 5 y^4)
    = DIV
    Limit does not exist because if I try to go through different paths, I get different limits.

    lim(x,y)->(0,3) ysin(x)/x
    = 3
    by separating the two functions y & sin(x)/x and evaluating them seperately

    lim(x,y)->(5,0) (3x + ln(1+xy) ) / (1+x+y)
    =15/6
    just by plugging in the values (function is continuous everywhere)

    lim(x,y)->(0,0) (x^2 * y^2) / (x^4 + 5y^4)
    = DIV
    Limit does not exist once again because of different paths I took = different limits

    Thanks a lot =)
     
  2. jcsd
  3. May 22, 2008 #2
    You're confusing how to find different paths. Instead, what you're doing is finding the limits at different points. Think of the limit as this:

    [tex]\lim_{\textbf{r}\rightarrow (0,0)}f(x,y)[/tex].

    Here, [tex]\textbf{r}[/tex] just represents some arbitrary path along the surface to that point.
     
  4. May 22, 2008 #3
    I understand what you mean but, for example, if I take the fourth limit:
    lim(x,y)->(0,0) (x^2 * y^2) / (x^4 + 5y^4)

    If I evaluate this when x=y then I end up with (x^4)/(6x^4) which gives a limit of 1/6.
    However, if I evaluate this at (x,0) which should also work, I get a limit of 0/(x^4) = 0
    Because both these limits (along different paths) are different, then the entire limit can not possibly converge.
     
  5. May 22, 2008 #4
    from what I can see, you're correct. You sure you typed it in right?
     
  6. May 22, 2008 #5
    yeah, Im sure.. but to get it right they all have to be right.. Im pretty confident about the last two limits.. its mainly the first two that are bugging me
     
  7. May 22, 2008 #6

    Defennder

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    Along which paths did you take the limits to evaluate the first one?
     
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