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Having problem solving a limit

  1. Oct 13, 2011 #1
    having problem solving a limit!!

    1. The problem statement, all variables and given/known data

    Solve the following limits algebraically.

    lim x->5 x^2 - 25 / √(2x+6) -4

    2. Relevant equations



    3. The attempt at a solution

    i've tried a different few ways.

    so by factoring the top i have (x+5)(x-5) so i get x=5 or x=-5, and while 5 is the limit, -5 on top with the square it will just render the negative sign useless.

    tried to multiply the bottom cognitive
    (x^2-25)(√(2x+6)+4)/(√(2x+6)-4)(√(2x+6)+4) =
    =(x^2-25)(√(2x+6)+4)/(√(2x+6))^2 - (4)^2 =
    =(x^2-25)(√(2x+6)+4)/2x+6-16=
    =(x^2-25)(√(2x+6)+4)/2x-10

    i'm stuck here and i doubt this is even the proper route.
     
  2. jcsd
  3. Oct 13, 2011 #2

    Mark44

    Staff: Mentor

    Re: having problem solving a limit!!

    Multiplying by the conjugate over itself is a useful approach.

    [tex]\lim_{x \to 5}\frac{(x-5)(x+5)}{\sqrt{2x+6}-4}\cdot \frac{\sqrt{2x+6} + 4}{\sqrt{2x+6} + 4}[/tex]
    [tex]= \lim_{x \to 5}\frac{(x-5)(x+5)(\sqrt{2x+6} + 4)}{(\sqrt{2x+6}-4) \cdot (\sqrt{2x+6} + 4)}[/tex]
    Can you take it from here?
     
  4. Oct 13, 2011 #3

    symbolipoint

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    Re: having problem solving a limit!!

    What you really mean is
    [tex]Lim_{x\rightarrow 5} \frac{x^2-25}{\sqrt{2x+6}-4}[/tex]

    ...

    Most of your work looks good. Try then dividing numerator and denominator by either x, or x^2 ?
     
  5. Oct 13, 2011 #4

    symbolipoint

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    Re: having problem solving a limit!!

    He appeared to be doing that; maybe I was confused looking at all the steps and symbols through text?
    EDIT: Give me the chance to try the problem myself; maybe more comment later.

    OK, very neat. Just multiplying numerator and denominator by conjugate gives you something to simplify and then simple substitution of x=5 can be evaluated with no complications.
     
    Last edited by a moderator: Oct 13, 2011
  6. Oct 13, 2011 #5

    HallsofIvy

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    Re: having problem solving a limit!!

    Both numerator and denominator are 0 when x= 5 therefore both have a factor of x- 5:
    [tex]\frac{(x- 5)(x+ 5)(\sqrt{2x+ 6}+ 4)}{2(x- 5)}[/tex]
     
  7. Oct 13, 2011 #6
    Re: having problem solving a limit!!

    thank you guys, when I woke up this morning and having looked at a few other problems I arrived at the same solution as HallsofIvy, I then removed the (x-5) and substituted the x for 5 and i got 40 as the limit I think.
     
  8. Oct 13, 2011 #7

    symbolipoint

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    Re: having problem solving a limit!!

    Yes, and in fact, you both think and KNOW for sure now.
     
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