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Homework Help: Limit of a Trigonometric Function

  1. Oct 9, 2005 #1
    Question:

    lim(x->0) for (tanx - sinx) / (sinx)^2

    This is what I got:

    = (sinx-sinxcosx) / (cosx)(sinx)^2
    = (sinx)(1-cosx) / (sinx)(sinx)(cosx)
    = (1 - cosx) / (sinx)(cosx)

    However, I can't figure out what to do from this step, as the limit still equals 0/0 at this stage.
     
  2. jcsd
  3. Oct 9, 2005 #2

    mezarashi

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    Homework Helper

    A 0/0 answer is the right prerequisite for using L'Hopital's rule.

    That is, the limit of any such function f(x) = h(x)/g(x), is h'(x)/g'(x).
    Try differentiating the top and bottom separately and plug in the numbers again.
     
  4. Oct 9, 2005 #3
    I haven't learned that yet.

    We are expected to solve the problem with using only limit laws and the fact that the lim (x->0) for sinx / x equals 1.
     
  5. Oct 9, 2005 #4
    ??? .
     
  6. Oct 9, 2005 #5

    mezarashi

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    Homework Helper

    Then you need to approach the problem differently. It is purely an algebriac/trigonometric problem. The strategy is to rid the denominator of any possible 0 terms (i.e. sin x).

    Edit - Here, try this:
    [tex] \frac{\tan x - \sin x}{sin^2x}[/tex]

    [tex]\frac{\frac{\sin x - \sin x \cos x}{\cos x}}{1 - \cos^2 x}[/tex]

    Keep in mind:
    [tex](1-a^2) = (1-a)(1+a)[/tex]
     
    Last edited: Oct 9, 2005
  7. Oct 10, 2005 #6

    HallsofIvy

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    Science Advisor

    Do you know that [tex]lim_{x\rightarrow0}\frac{sin x}{x}= 1[/tex]?
    Do you know that [tex]lim_{x\rightarrow0}\frac{1- cos x}{x}= 0[/tex]?

    Can you figure out how to write [tex]\frac{1-cos x}{sin x}[/tex] in terms of [tex]\frac{sin x}{x}[/tex] and [tex]\frac{1- cos x}{x}[/tex]?
     
  8. Oct 10, 2005 #7
    (1 - cosx) / (sinx)(cosx) *
    (1+cosx) / (1+cosx) = ...

    Or:
    (tanx - sinx) / (sinx)^2 =
    tanx (1 - cosx) / (1 - (cosx)^2) = ...
    (look up Mezarashi's hint)
     
    Last edited: Oct 11, 2005
  9. Nov 20, 2011 #8
    lim x->0 of (tanx - sinx)/(sinx)^2

    lim x->0 of (tanx)/(sinx)^2 - (sinx)/(sinx)^2

    lim x->0 of (sinx)/(cosx(sinx)^2) - (1/sinx)

    lim x->0 (1/cosxsinx) - (1/sinx)

    lim x->0 (1-cosx)/(cosxsinx)

    lim x->0 (-(cosx-1)/x) / (cosxsinx)/(x))

    lim x->0 (-0)/ (1(1)) = 0

    so the final answer is 0
     
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