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Limit of Trigonometric f'n as x approaches 0

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the following limit as x tends to zero
    (idk how to code this stuff properly so the lim is lim as x->0)
    lim (sinx-tanx)/x^3


    2. Relevant equations

    lim(sinx)/x = 1
    lim(cosx) = 1
    lim(1-cosx)/x = 0

    3. The attempt at a solution

    attempted to multiply by conjugate
    attempted to remove the root from the top and the bottom
    no idea how to do this.
     
  2. jcsd
  3. Nov 3, 2011 #2
    I'm not sure what you are allowed to do here, but one way to solve this limit would be to consider the Taylor series of sin and tan about the point x = 0.

    In particular, [itex]\sin(x) \approx x - x^3/3! + x^5/5! - ...[/itex] and [itex]\tan(x) \approx x + x^3/3 + 2x^5/5 + ...[/itex]. Hence, subtracting these two series, gives:

    [tex]\sin(x) - \tan(x) \approx -(x^3/3! + x^3/3) + O(x^5)[/tex]

    Substituting this into your function (i.e. dividing the above by [itex]x^3[/itex]) and taking the limit as that goes to zero will give you an answer.
     
  4. Nov 3, 2011 #3
    nope. it has to be done without L'Hospital's rule, and we learn the taylor / maclaurin series next semester?. My attempts to rearrange it only get down to x on the bottom, can't seem to shake it out.
     
  5. Nov 3, 2011 #4
    Okay... here is a slightly more cumbersome way,

    Notice,
    [tex]\frac{\sin(x) - \tan(x)}{x^3} = -\left(\frac{\sin(x)}{x}\right)^3\frac{1}{\cos^2(x) + \cos(x)}[/tex].

    A simple application of limit laws will give you the solution from that stage.

    Hint:
    To obtain the above equality, start by multiplying the top and bottom of your fraction by [itex](\sin(x)+\tan(x))[/itex].
     
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