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Homework Help: Limits with sine and square roots

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    solve the limit: [tex]\lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}[/tex]




    3. The attempt at a solution

    Is there a way of how i can solve this without using l'hospital's rule or taylor series?
     
  2. jcsd
  3. Feb 23, 2012 #2

    Ray Vickson

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    What is wrong with using l'Hospital's rule or Taylor series? Of course, changing to the variable [itex] t = \sqrt{x}[/itex] simplifies the question a lot.

    RGV
     
    Last edited: Feb 23, 2012
  4. Feb 23, 2012 #3
    Well i haven't learned l'hospitals rule or taylor's series so i can't use either of them.
     
  5. Feb 23, 2012 #4
    Multiply numerator and denominator by [itex]\sqrt{x}+1[/itex].
     
  6. Feb 23, 2012 #5
    ok i multiplied by √x+1 and i got:

    [tex]\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}[/tex]

    now should i cancel the x-1?
     
  7. Feb 23, 2012 #6
    Do you know the limit

    [tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}[/tex]
     
  8. Feb 23, 2012 #7
    yes it's 1
     
  9. Feb 23, 2012 #8
    Now, use that to calculate the limit.
     
  10. Feb 23, 2012 #9
    alright got it now:

    [tex]\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2[/tex]

    Is that correct?
     
  11. Feb 23, 2012 #10

    Mark44

    Staff: Mentor

    No.
    [tex]\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1[/tex]

    Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
    [tex]\lim_{y \to 0}\frac{sin(y)}{y} = 1[/tex]
     
  12. Feb 23, 2012 #11
    We still haven't learned how to split a limit into 2 different limits yet.
     
  13. Feb 23, 2012 #12

    Mark44

    Staff: Mentor

    Are you sure? It's a very basic property of limits, and one that is presented pretty early.
    [tex]\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)[/tex]

    The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.
     
  14. Feb 23, 2012 #13
    Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.
     
  15. Feb 23, 2012 #14
    Then I'm afraid you can't solve this question.

    You can't even solve questions like [itex]\lim_{x\rightarrow 1}{x^2}[/itex] in this case.
     
  16. Feb 23, 2012 #15

    SammyS

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    So, if [itex]\displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\,,[/itex] then what is [itex]\displaystyle \lim_{x-1\rightarrow 0}\frac{\sin(x-1)}{x-1}\,?[/itex]
     
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