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Find the limit of the this equation

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limit if it exists.


    2. Relevant equations
    [tex]lim_{x\to2} \frac{\sqrt{x-6}-2}{\sqrt{x-3}-1}[/tex]


    3. The attempt at a solution
    Completely lost, I've tried, finding the conjugate of both the numerator or denominator and I'm unable to find a solution.
     
  2. jcsd
  3. Mar 5, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you sure that is the problem? AT x= 2, the fraction is
    [tex]\frac{2i- 2}{i- 1}= 2[/tex]
    Nothing hard about that!

    Perhaps you meant
    [tex]\frac{\sqrt{6- x}- 2}{\sqrt{3- x}-1}[/itex]
    which is "0/0" when x= 2.
    Use the "conjugates" as you mention- multiply both numerator and denominator by [itex](\sqrt{x-6}+2)(\sqrt{x-3}+1)[/itex]. You will get x2-4 in the numerator and x-2 in the denominator and can cancel.
     
  4. Mar 5, 2007 #3
    ok, here's another one I don't understand.

    2. Relevant equations
    [tex]lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}[/tex]

    3. The attempt at a solution
    [tex]lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}(\frac{x-1}{x-1})[/tex]
    [tex]lim_{t \to 0} \frac{(x^2-1)}{(x^2-1)sin(\Pi x)}[/tex]
    [tex]lim_{t \to 0} \frac{(1)}{sin(\Pi x)}[/tex]

    At this point I get stuck... the [tex]sin(\Pi x)[/tex] is still 0. I tried using some of the trig identities, that I actually remembered, and got this.

    [tex]lim_{t \to 0} \frac{(sin^2(\Pi x) + cos^2(\Pi x)}{sin(\Pi x)}[/tex]

    Now I thought if I could get I could get the numerator in terms of sine, I could simple divide into it, but I'm not sure if this would work :-/... any suggestions?
     
    Last edited: Mar 5, 2007
  5. Mar 5, 2007 #4
    is it possible for me to do

    [tex]lim_{t \to 0} \frac{(1)}{sin(\Pi x)}[/tex]
    [tex]lim_{t \to 0} \frac{(1)}{csc(\Pi x)}[/tex]
    [tex]lim_{t \to 0} sin(\Pi x)[/tex]

    ?
     
  6. Mar 5, 2007 #5
    mmm never mind. there was a problem in my solution. [tex]xsin(\Pi x)(x-1)[/tex] does not equal [tex](x2-1)sin(\Pi x)[/tex]... I will try again...
     
  7. Mar 5, 2007 #6
    ok, I think I got it.

    [tex]lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}[/tex]

    [tex]lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}(\frac{csc(\Pi x)}{csc(\Pi x)})[/tex]

    [tex]lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}[/tex]

    [tex]lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}[/tex]

    [tex]lim_{t \to 1} csc(\Pi x) + \frac{(csc(\Pi x)}{x}[/tex]

    [tex]lim_{t \to 1} csc(\Pi * 1) + \frac{(csc(\Pi *1)}{1}[/tex]

    [tex]lim_{t \to 1} csc(\Pi) + (csc(\Pi)[/tex]

    [tex]lim_{t \to 1} 2csc(\Pi)[/tex]

    [tex]lim_{t \to 1+} 2csc(\Pi) = +\infty[/tex]

    [tex]lim_{t \to 1-} 2csc(\Pi) = -\infty[/tex]
     
    Last edited: Mar 6, 2007
  8. Mar 5, 2007 #7
    Sorry, the limit is t->1+ not t->0...
     
  9. Mar 5, 2007 #8
    ok, now if I take the left hand limit and right hand limit, the right hand limit approaches infinity while the left hand appraches negative infinity?
     
  10. Mar 6, 2007 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Does it bother you at all that you are taking the limit as t-> 1 but there is no t in the formula?! (Mathematics requires precision- be careful what you write!)

    Perhaps you meant
    [tex]lim_{t \to 1^+} \frac{t+1}{tsin(\Pi t)}[/tex]

    I don't see difficulty with that. The denominator is continuous and at t= 1 is [itex](1)sin(\pi)= 0[/itex] but the numerator does NOT go to 0. What does that tell you?

    (Don't capitalize "Pi"- you want \pi: [itex]\pi[/itex].)
     
  11. Mar 6, 2007 #10
    o_O... I am such a dork...

    [tex]\lim_{t \to 1^+} \frac{t+1}{tsin(\pi t)} = -\infty [/tex] as [tex]t \to 1^+[/tex]
     
    Last edited: Mar 6, 2007
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