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Lim (sinx-cosx)/(pi-4x) where x->pi/4

  1. Jan 6, 2010 #1
    lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Hi! Could someone help me to solve this problem without using L'Hospital rule? Any hint would be splendid...

    lim (sinx-cosx)/(pi-4x)
    x-->pi/4

    Thanks in advance!
     
  2. jcsd
  3. Jan 6, 2010 #2

    ideasrule

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    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    L'Hopital's rule?
     
  4. Jan 6, 2010 #3
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Sorry, but without L'Hospital. I forgot to mention...With L'Hospital is easy to solve and it gives
    =-sqrt(2)/4
     
  5. Jan 6, 2010 #4

    rock.freak667

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    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Try rewriting sin(x)-cos(x) in the form Rsin(x-A)
     
  6. Jan 7, 2010 #5
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    You could multiply by (cosx + sinx)/(cosx + sinx) and use the identity cos2x -sin2x = cos2x. Then let u = 2x, use the identity sin([itex]\pi[/itex]/2 - x) = cosx, rewrite it some more, and use the common trig limit sinx/x. Don't forget to change the limits with each substitution.
     
  7. Jan 7, 2010 #6
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Wow! Thank you for this great tip! I tried a similar method but I had problems to find a good substitution. u=2x really works and the problem is solved!!! Thanks again!
     
  8. Jan 7, 2010 #7
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Now I have two solutions for my limit problem. I figured out how to transform sin(x)-cos(x) into Rsin(x-A) form:

    R=Sqrt(2)
    A=pi/4

    sinx-cosx=Sqrt(2)*sin(x-pi/4)

    Thank you for your help!
     
  9. Jan 9, 2010 #8
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    L'Hopital's rule is simple. It says that if the limit of a quotient takes on an indeterminate form, such as 0/0 then the limit is equivalent to the derivative of the numerator divided by the derivative of the denominator.

    sinx-cosx / pi -4x

    set the same limit to

    cosx+sinx / -4
     
  10. Jan 9, 2010 #9
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Alternatively, you could expand [itex]\sin(x)[/itex] and [itex]\sin(x)[/itex] in a Taylor's Series about [itex]x = \pi/4[/itex]. The numerator subtraction will eliminate even power terms and the denomiator will cancel one power of [itex](x-\pi/4)[/itex] term. The result will have a leading constant term which is your limit as [itex]x\rightarrow \pi/4[/itex].
     
  11. Jan 9, 2010 #10
    Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

    Hmm, I'm not familiar with expanding in a Taylor's Series.
     
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