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Finding a limit R²->R

  1. Sep 11, 2010 #1
    Hi there. Well, in the next exercise I must find the limit of [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{\cos(xy)-1}{x}}[/tex], if it exists. I wanna know if I did it right.

    If [tex]y=cx[/tex]

    [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{\cos(xy)-1}{x}}=\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{\cos(cx^2)-1}{x}}=\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{-\sin(cx^2)2cx-1}{1}}=-1[/tex]

    If [tex]x=cy[/tex]

    [tex]\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{\cos(cy^2)-1}{cy}}=\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{-\sin(cy^2)2cy-1}{cy}}=-\displaystyle\frac{1}{c}[/tex]

    [tex]\therefore{\not{\exists}}\textsf{double limit}[/tex]

    So, what you say?

    By there, and thanks for posting.

    PD: Ok, Now I see, after plotting with wolfram, some calculus errors I've committed. The limit actually seems exists. So I should use the delta epsilon definition of limits to make a demonstration.
    Last edited: Sep 11, 2010
  2. jcsd
  3. Sep 11, 2010 #2


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    If y=cx, then x=dy where d=1/c. You shouldn't get different answers trying each variety.

    It looks like you're trying L'hopital's rule, but you aren't differentiating properly (the derivative of 1 is zero for example, and the derivative of cy is not going to be cy). Furthermore you're only differentiating with respect to one variablei

    Try expanding cos(xy) in terms of a power series
  4. Sep 11, 2010 #3
    You mean like using Taylor? I don't know how to with two variables. I see I've committed a few mistakes. As you say, it should give the same limit, which is zero. But, to demonstrate that zero is the limit I must use the delta epsilon definition, right?
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