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Homework Help: Solving limit with algebraic tools

  1. Nov 25, 2005 #1
    Hello!
    [tex]
    \lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}
    [/tex]
    I solved this limit using L'Hopital and expanding trigonometric functions to series.
    But i have to solve it using algebraic tools (without series). I don't know how to do it. [tex]\sin x - x[/tex] looks difficult to deal with.
     
    Last edited: Nov 25, 2005
  2. jcsd
  3. Nov 25, 2005 #2

    benorin

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    You have likely proved that

    [tex]\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1[/tex] (EDIT: yes, I meant 0, not [itex]\infty[/itex], thanks)

    If so, use it.
     
    Last edited: Nov 26, 2005
  4. Nov 25, 2005 #3

    VietDao29

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    ???
    :confused:
    Are you sure the limit above is correct? Let's check it again. :wink:
    Shouldn't it read:
    [tex]\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1[/tex]
    Or
    [tex]\lim_{x\rightarrow \infty} \frac{\sin(x)}{x}=0[/tex]
     
  5. Nov 26, 2005 #4
    Possibly, Benorin meant [tex]\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1[/tex],
    but i believe, that this can't help in solving initial problem. I have to use further terms of sinx series. [tex]\sin x = x - \frac{x^3}{3!}+...[/tex] But how can i get them algebraically??
     
  6. Nov 28, 2005 #5
    Realy, nobody have any idea about initial problem?
     
  7. Nov 28, 2005 #6

    matt grime

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    there aren't many things that spring to mind except using your trig identities.
     
  8. Nov 28, 2005 #7
    So, i tried to get some more terms of [tex]\sin x [/tex] expansion. Here is my result:
    [tex]
    \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} = 2 \sin \frac{x}{2} (\cos \frac{x}{2} -1 +1) = -2 \sin \frac{x}{2} (1-\cos \frac{x}{2}) + 2\sin \frac{x}{2} = -4 \sin \frac{x}{2} (\sin \frac{x}{4})^2 +2\sin \frac{x}{2}
    [/tex]
    Now if x->0 i got that [tex]\sin x = x - \frac{x^3}{8}+...[/tex] and not [tex]\sin x = x - \frac{x^3}{3!}+...[/tex] as it should be. Where do i went wrong?
     
    Last edited: Nov 28, 2005
  9. Nov 29, 2005 #8

    VietDao29

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    If you drop off all of the terms of power higher than 1 in 2 sin(x / 2), you must as well drop off all the term has the poer higher than 1 in the other terms.
    [tex]-4\sin \frac{x}{2} \sin ^ 2 \frac{x}{4} = -\frac{x ^ 3}{8}[/tex] should be also dropped off, as the degree of x is 3 (which is obviously greater than 1).
    That'll give:
    [tex]\sin x = x + ...[/tex], and that's correct.
    --------------------
    This is my approach:
    I'll prove
    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x > x - \frac{x ^ 3}{3!}[/tex]
    Or slightly differently:
    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x - x + \frac{x ^ 3}{3!} > 0[/tex]
    Let f(x) := sin x - x + x3 / 3!
    f'(x) = cos x - 1 + x2 / 2!
    f''(x) = -sin x + x
    f'''(x) = -cos x + 1
    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : f'''(x) > 0[/tex] that means f''(x) is increasing on that interval, f''(0) = 0 => [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : f''(x) > 0[/tex], that again means f'(x) is increasing on that interval, f'(0) = 0 => [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : f'(x) > 0[/tex].
    That means f(x) is increasing on that interval, f(0) = 0, so we have:
    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : f(x) > 0[/tex] (Q.E.D)
    -------------------
    You can do exactly the same to prove:
    [tex]\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x < x - \frac{x ^ 3}{3!}[/tex]
    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x < x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!}[/tex]
    [tex]\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x > x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!}[/tex]
    -------------------
    It means:

    [tex]\forall x \in ]0, \frac{\pi}{2} [ \ : -\frac{x ^ 3}{3!} \leq \sin x - x \leq -\frac{x ^ 3}{3} + \frac{x ^ 5}{5!}[/tex]
    [tex]\forall x \in ]-\frac{\pi}{2}, 0 [ \ : -\frac{x ^ 3}{3!} \geq \sin x - x \geq -\frac{x ^ 3}{3} + \frac{x ^ 5}{5!}[/tex]
    From there, just use Squeeze theorem, to evaluate the limit of:
    [tex]\lim_{x \rightarrow 0} \frac{\sin x - x}{x ^ 3}[/tex].
    And, that's way longer than to use L'Hopital rules, or sin expansion. Maybe someone will come up with something shorter.
     
    Last edited: Nov 29, 2005
  10. Nov 29, 2005 #9
    Thank you very much, VietDao. Your approach is realy very nice. But unfortunately i cannot use derivatives :cry: . Just algebra:yuck:
     
  11. Nov 29, 2005 #10

    NateTG

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    Try
    [tex]\frac{-x(1 -\cos x)}{\sin x - x} \times \frac{1+\cos x}{1+\cos x}[/tex]
    and some trig identities.
     
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