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Homework Help: Trying to solve a limit

  1. Feb 23, 2006 #1
    Hi
    I've been trying to solve a limit, and I found some difficulties doing that. Any help would be appreciated!


    lim cos(2x) / 2cosx - √2
    x→ (π/4)

    so, what I tried to do is expand that cos(2x) but I still get 0/0 :frown:
    Any hints ?
    Thank cyou
     
  2. jcsd
  3. Feb 23, 2006 #2

    arildno

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    First, get your parentheses right!

    Secondly, have you learnt L'Hopital's rule yet?
     
  4. Feb 23, 2006 #3
    Sorry... but I don't see anything wrong with them!
    No we have not learned it yet, and I would like to learn it.
     
  5. Feb 23, 2006 #4

    arildno

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    Oh, really?
    What you've ACTUALLY written means:
    [tex]\frac{\cos(2x)}{2\cos(x)}-\sqrt{2}[/tex]

    But that wasn't the main point, though.
     
  6. Feb 23, 2006 #5
    That's right, sorry!
     
  7. Feb 23, 2006 #6

    arildno

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    In order to proceed without L'Hopital's rule, multiply with the conjugate expression:
    [tex]\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}*\frac{2\cos(x)+\sqrt{2}}{2\cos(x)+\sqrt{2}}=\frac{\cos(2x)}{4\cos^{2}(x)-2}(2\cos(x)+\sqrt{2})=\frac{\cos(2x)}{2\cos^{2}(x)-1}\frac{2\cos(x)+\sqrt{2}}{2}[/tex]
    Now, any thoughts of simplifying this further?
     
  8. Feb 23, 2006 #7
    I can see that (cos(2x))/(2cos²(x)-1) equals to 1.
    Does that mean, the limit is equal to sqrt(2) ?
    If that's right, how can it be solved by that L'Hopital rule ?
     
  9. Feb 23, 2006 #8

    arildno

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    That's correct!
    Now, to introduce L'Hopital's rule, lets consider h(x)=f(x)/g(x), and search for h(X) when f(X)=g(X)=0 (X being the point we let x approach.

    Now, according to the mean value theorem, there exists an y in the interval (X,x) so that f(x)=f(X)+f'(y)(x-X) and similarly, a z in (X,x), so that g(x)=g(X)+g'(z)(x-X) (y and z need not be the same number!!)
    Thus, we have:
    [tex]h(x)=\frac{f(X)+f'(y)(x-X)}{g(X)+g'(z)(x-X)}=\frac{f'(y)}{g'(z)},y,z\in(X,x)[/tex]
    since f(X)=g(X)=0, and [itex]X\neq{x}[/itex]

    Now, as you let x approach X, then evidently we must have that y approaches X, and z approaches X.
    Thus, you get:
    [tex]\lim_{x\to{X}}h(x)=\frac{f'(X)}{g'(X)}[/itex]
    insofar as this is defined.

    Thus, L'Hopital's rule says that if h(X) is a 0/0-expression, you can find its limit value by computing the ratio between the the derivatives of f and g, evaluated at X.

    In your example:
    [tex]\lim_{x\to\frac{\pi}{4}}\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\lim_{y,z\to\frac{\pi}{4}}\frac{-2\sin(2y)}{-2\sin(z)}=\frac{\sin(2\frac{\pi}{4})}{\sin(\frac{\pi}{4})}=\sqrt{2}[/tex]
     
  10. Feb 23, 2006 #9
    Ah I understood a little bit... because we haven't learned derivatives yet :-/
    Thanks anyway!
     
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