1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    ???
    :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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Try
    [tex]\frac{-x(1 -\cos x)}{\sin x - x} \times \frac{1+\cos x}{1+\cos x}[/tex]
    and some trig identities.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solving limit with algebraic tools
  1. Algebra of Limits (Replies: 1)

  2. Solve the limit (Replies: 2)

Loading...