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Limit with trig function

  1. Dec 6, 2004 #1
    I am completly lost on where to start, any help would be appreciated:

    lim (x->0) tan^2(2x) / 3x^2

    I do not have any work so far, but I'm trying to find a way to get rid of the x^2 in the denominator, which so far has been fruitless.
  2. jcsd
  3. Dec 6, 2004 #2


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    Try replacing tan(2x) with sin(2x)/cos(2x). You probably know something about the limit of sin(x)/x as x->0.
  4. Dec 6, 2004 #3


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    This limit works fine without the aid of Mr.L'Ho^pital.
    Are u 100% convinced that:
    [tex] \lim_{2x \rightarrow 0} \frac{\tan{2x}}{2x} =1 [/tex]

    If so,u'll be able to apply the formula above properly and get the result.

    One more hint:
    [tex]\lim_{x \rightarrow 0} \frac{\tan^{2} 2x}{3x^2} = \frac{4}{3}
    (\lim_{2x \rightarrow 0} \frac{\tan{2x}}{2x})^{2} [/tex]

    I hope you can take it from there.
  5. Dec 7, 2004 #4
    Thank you both, I have my answer now.
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