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Two Limits Questions

  1. Jul 16, 2006 #1
    Hi, I have think of the followed two questions quite a long time,

    a)
    [tex]\lim_{x\rightarrow 0+} (\sqrt{\frac{3}{x}+\frac{4}{x^2}+5} - \sqrt{\frac{2}{x}+\frac{4}{x^2}+6})[/tex]

    b)
    [tex]\lim_{x\rightarrow 2} (\frac{|x-3|-|3x-5|}{x^2-5x+6}) [/tex]


    How to solve this two questions?
     
  2. jcsd
  3. Jul 16, 2006 #2
    What have you done so far to solve the problems ?

    1) Multiplying by the conjugate would help.
    2) Try breaking the mudulus signs,
     
  4. Jul 16, 2006 #3

    quasar987

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    a) is an indeterminate form of the type [itex]\infty - \infty[/itex]

    "indeterminate form" means you cannot conclude to the value of the limit. It could be anything. So you have to put it under a different form, where you can conclude. What have you tried

    b) This is an indeterminate form of the type 0/0. you can apply l'Hospital.
     
  5. Jul 16, 2006 #4

    quasar987

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    Forget what I said for b). arunbg's advice is much better.
     
  6. Jul 16, 2006 #5

    matt grime

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    mutlipliying, as it stands by the conjugate won't help, since the conjugate is also not defined. First try pulling out a 1/x^2, and then try using the difference of two squares thing.
     
  7. Jul 16, 2006 #6
    Hi,

    My answer is here:
    a)
    http://www.geocities.com/myjunkmail31/Limit1.jpg
    b)
    http://www.geocities.com/myjunkmail31/Limit2.jpg
    is it correct?(Thanks for the hint=))

    Another way, if the question is
    [tex]\lim_{x\rightarrow 0-} (\sqrt{\frac{3}{x}+\frac{4}{x^2}+5} - \sqrt{\frac{2}{x}+\frac{4}{x^2}+6})[/tex]
    PS:Original question is 0+, here is 0-

    From the computer generated graph, it seems to have a negative value, but how I get it?
    Since using the method here
    http://www.geocities.com/myjunkmail31/Limit2.jpg
    No matter it is 0+ or 0-, I will always get 1/4.
     
  8. Jul 16, 2006 #7

    VietDao29

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    Both of your answers look fine to me. :smile: Congratulations.
    However, the b limit can be also solved by breaking the absolute value, something like this:
    Since we have:
    [tex]|A| = \left\{ \begin{array}{ll} A, & \mbox{if A} \geq 0 \\ -A, & \mbox{if A} < 0 \end{arrray} \right.[/tex]
    [tex]\lim_{x \rightarrow 2} x - 3 = -1 < 0[/tex], and [tex]\lim_{x \rightarrow 2} 3x - 5 = 1 > 0[/tex]. So, we have:
    [tex]\lim_{x \rightarrow 2} \frac{|x - 3| - |3x - 5|}{x ^ 2 - 5x + 6} = \lim_{x \rightarrow 2} \frac{(3 - x) - (3x - 5)}{(x - 2)(x - 3)} = \lim_{x \rightarrow 2} \frac{-4x + 8}{(x - 2)(x - 3)} = -4 \lim_{x \rightarrow 2} \frac{x - 2}{(x - 2)(x - 3)} = -4 \lim_{x \rightarrow 2} \frac{1}{(x - 3)} = (-4) (-1) = 4[/tex]
    ---------------------
    For your third question, it's a one-sided limit, and x tends to 0 from the negative side; i.e, x should be negative, right?
    And, so in the first line, we have:
    [tex]\lim_{x \rightarrow 0 ^ -} \sqrt{\frac{3}{x} + \frac{4}{x ^ 2} + 5} - \sqrt{\frac{2}{x} + \frac{4}{x ^ 2} + 6} = \lim_{x \rightarrow 0 ^ -} \sqrt{\frac{1}{x ^ 2}} \times (\sqrt{5x ^ 2 + 3x + 4} - \sqrt{6x ^ 2 + 2x + 4}) = \lim_{x \rightarrow 0 ^ -} \left| \frac{1}{x} \right| \times (\sqrt{5x ^ 2 + 3x + 4} - \sqrt{6x ^ 2 + 2x + 4})[/tex]
    Since x < 0, breaking the absolute value, we have:
    [tex]= - \lim_{x \rightarrow 0 ^ -} \frac{1}{x} \times (\sqrt{5x ^ 2 + 3x + 4} - \sqrt{6x ^ 2 + 2x + 4}) = ...[/tex], and from here, you can just do exactly what you did in a.
    Can you get this? :)
     
    Last edited: Jul 16, 2006
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