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A limit :s

  1. Mar 2, 2006 #1
    hey again!
    this time, I have this one to solve

    [tex]\lim_{x\rightarrow 0}\frac{\tan(x)-\sin(x)}{x^3}[/tex]

    i went like this

    [tex]\lim_{x\rightarrow 0}\frac{\frac{tan(x)}{x} - \frac{sin(x)}{x}}{x^2}[/tex]

    = lim (0/0)

    which is always an undetermined form... is there any other way to solve this WITHOUT using derivatives (not learned yet)

    Thank you!
     
    Last edited: Mar 2, 2006
  2. jcsd
  3. Mar 2, 2006 #2

    arildno

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    There's ALWAYS another way! :smile:
    Here's how you could start out:
    [tex]\lim_{x\to[0}}\frac{\tan(x)-\sin(x)}{x^{3}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{\frac{1}{\cos(x)}-1}{x^{2}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{1-\cos^{2}(x)}{\cos(x)(1+\cos(x))x^{2}}=\lim_{x\to{0}}(\frac{\sin(x)}{x})^{3}\frac{1}{\cos(x)(1+\cos(x))}[/tex]
    Can you take it from there?
     
  4. Mar 2, 2006 #3
    I think so,
    So the answer is 1/2 ?
     
  5. Mar 2, 2006 #4

    arildno

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    You are not sure about that?
     
  6. Mar 2, 2006 #5
    I am actually!
    Thank you
     
  7. Mar 2, 2006 #6

    arildno

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    You're welcome.
     
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