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Solving infinity limit, something weird.

  1. Apr 1, 2007 #1
    1. The problem statement, all variables and given/known data
    This question asks you to evaluate the limit of the following function

    lim sqrt(9x^2-3x) -3x
    x-> infinity

    I did it using 2 methods:

    lim sqrt(9x^2-3x) -3x
    x-> infinity

    = lim x(sqrt(9-3/x)) -3x
    x-> infinity

    = lim 3x - 3x
    x -> infinity

    = 0

    Second Method:
    I multiplied by the conjugate radical/conjugate radical and after simplification, I get

    lim -3x/6x
    x -> infinity

    = 1/2

    Why is it that I'm getting different answers? Which solution is the right one? Also, how do I type up mathematical formulas? Can someone point me to a LaTex tutorial? Thanks!
  2. jcsd
  3. Apr 1, 2007 #2
    when you multiplied the conjugate, did you also multiply the denominator?
  4. Apr 1, 2007 #3
    your first attempt is wrong because your trying to solve the limit of only one part of your equation without the other.
    You can't solve the lim (x-> infinity) of sqrt(9-3x) without solving the rest.

    Your second answer is right except for the sign, it should be -1/2.
    Last edited: Apr 1, 2007
  5. Apr 1, 2007 #4
    Alright thanks to both of you, I have another question though. If I evaluate both parts in an equation and end up with

    3x - 3x
    Where x is approaching infinity, would it be right to say that the limit is 0? I remember my teacher saying that you cannot perform arithmetic operations involving infinity, then how come you can divide out -3x/6x? I'm new to calculus so please help lol.
  6. Apr 2, 2007 #5
    if it is 3x- 3x, then yes because 3[ lim x(1-1)]=0
    Last edited: Apr 2, 2007
  7. Apr 2, 2007 #6
    my bad, what I had on there earlier was a mistake, sorry. Infinity multiplied by zero should equal zero right? So it should approach zero, I think. Can anybody explain clearly the properties of infinity for limits and its relationship with the concept of zero? Hey wk1989, you are asking very intelligent questions, and I felt the same frustration when I took Calc I. They slap me with these concepts and I was to take it in without much clear explanations that I could imagine. If you are going to take Calc. II, when you study series, this kind of stuff will be explained further, so don't worry.
    Last edited: Apr 2, 2007
  8. Apr 2, 2007 #7
    noooooo, infinity multiplied by zero is undefined
  9. Apr 2, 2007 #8
    why? Can you explain the reasoning behind it?
  10. Apr 2, 2007 #9
    i guess what u really meant is this, couse u forgot a pair of brackets here
    lim sqrt(9x^2-3x) -3x
    x-> infinity
    and this

    lim (sqrt(9x^2-3x) -3x), x->infinity

    are not the same.
    however the answer like the others stated is -(1/2)
  11. Apr 2, 2007 #10


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    Infinity is not a number (i.e. it is not a member of the real, complex, rational, etc.. numbers) and so multiplication by infinity is undefined.
  12. Apr 2, 2007 #11


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    That's a little pedantic isn't it? :rolleyes:
  13. Apr 2, 2007 #12
    you are saying that

    0*infinity =0, then if this would be right then this also wuld be right

    0/0=infinity, if we devide by zero your "equation". but we come to a contradiction here couse we know that 0/0 is undefined. or it also would be right according to what you are saying to conclude that

    (0/0)*infinity=0/0, which again is undefined.

    somebody else migh give a more effective and detailed reasoning i guess.
  14. Apr 2, 2007 #13
    no i think it's about writting it correctly.
  15. Apr 2, 2007 #14


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    In all fairness though, he wrote this [tex]\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}-3x[/tex], and you wrote this [tex]\lim(\sqrt{9x^2-3x}-3x), x\rightarrow\infty[/tex].

    The former is a more standard way of writing it!
  16. Apr 2, 2007 #15
    Yeah, that is logical mathematically
  17. Apr 2, 2007 #16
    i am not talking about what is standard and not, but you have to understand that these two expressions are not equivalent, they represent different things. At the former 3x is out of limit when x-> infinity, so it means that 3x at this case is absolutely not affected when x-> infinity, but at the latter one when 3x is inside the breackets it means that we should also consider 3x when we evaluate the limit of the whole expression when x-> infinity.

    you cristo are saying this in a more general form

    lim f(x)-g(x),x->infinity is equal to lim[f(x)-g(x)],x->infinity, which is absurd. because the latter one using the features of additon and subtraction of limits we get

    lim[f(x)-g(x)],x->infinity=lim f(x)-lim g(x),x->infinity, so we get something which is not the same as lim f(x)-g(x),x->infinity , as you are claiming to be. Once more these two limits are not the same.

    can someone else coment on this???
    Last edited: Apr 2, 2007
  18. Apr 2, 2007 #17


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    Unfortunately, you made this ambiguous by also moving the "[itex]x\rightarrow \infty[/itex]"!

    I understand that your point is that
    [tex]\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x})- 3x[/tex]
    is not the same as
    [tex]\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x}- 3x)[/tex]
    But I am not sure that I agree that one must interpret
    [tex]\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}- 3x[/tex]
    as the former rather than the latter.
  19. Apr 2, 2007 #18
    I think in order to avoid any misinterpretation or ambiguity, we should write it as i suggested. The way cristo said can also be written like this

    lim sqrt(9x^2-3x)-3x=-3x+lim sqrt(9x^2-3x), when x->infinity in both cases. however if we write it this way lim [sqrt(9x^2-3x)-3x] is not equal to -3x+lim sqrt(9x^2-3x), when x->infinity, in both cases again. I am sorry for the symbols.

    so this is my poing
    Last edited: Apr 2, 2007
  20. Apr 2, 2007 #19


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    Oh, I agree with that! The more parentheses the better!
  21. Apr 2, 2007 #20


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    Ok, standard was the wrong word to use, but I don't "have to understand they are not equivalent!" Given this limit [tex]\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}-3x[/tex], I (for one) would automatically interpret this as [tex]\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x}-3x)[/tex]. If one wanted to take the limit of only the first term, then one would put parentheses around only the first term. That was my point, and that's why I said it was a little pedantic to correct something that didn't need correcting!
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