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Calculating a limit with tangent as denominator

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data
    lim x^3-2x^2+x/tanx
    x->0


    3. The attempt at a solution

    All i know is that tan is going to break up into sinx/cosx so the equation will look like this

    lim x^3-2x^2+x/(sinx/cosx)
    x->0

    I haven't worked with cubic or quadratic functions yet so I don't know where to begin with this.
     
  2. jcsd
  3. Feb 16, 2012 #2

    Mentallic

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    Homework Helper

    Remember to throw in brackets where necessary, because the way it looks right now is

    [tex]\lim_{x\to 0}\left(x^3-2x^2+\frac{x}{\tan(x)}\right)[/tex]

    So you need to put brackets around the cubic to avoid this confusion :wink:


    If your problem is of the form [tex]\lim_{x\to a}\left(f(x)\cdot g(x)\right)[/tex] then this is equivalent to [tex]\lim_{x\to a}f(x)\cdot\lim_{x\to a} g(x)[/tex]

    So try factoring out an x from the numerator, and see if you can apply this idea correctly.
     
  4. Feb 16, 2012 #3
    I may have confused you with the way I wrote it, it should look like this. (I'll use brackets to make it clearer)

    lim (x^3-2x^2+x)/(sinx/cosx)
    x->0
     
  5. Feb 16, 2012 #4
    The answer eventually = 1 but I don't know how to get there
     
  6. Feb 17, 2012 #5

    ehild

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    Homework Helper
    Gold Member

    Factor out x as Mentallic suggested and rearange the expression as

    [tex]\left(\frac{x}{\sin(x)}\right)(x^2-2x+1)\cos(x)[/tex]

    ehild
     
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