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Math Help - Limit

  1. #1
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    Limit

    hi , help please
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  2. #2
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    Re: Limit

    What have you tried? Can you use L'Hopital's rule?

    -Dan
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  3. #3
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    Re: Limit

    Quote Originally Posted by abualabed View Post
    hi , help please
    If you're allowed to use L'Hospital's Rule, use it.

    If not, rewrite it as

    \displaystyle \begin{align*} \frac{\tan{(x)} - \sin{(x)}}{x^3} &= \frac{\tan{(x)}}{x^3} - \frac{\sin{(x)}}{x^3} \\ &= \frac{\sin{(x)}}{x^3\cos{(x)}} - \frac{\sin{(x)}}{x^3} \\ &= \frac{\sin{(x)}}{x} \cdot \frac{1}{\cos{(x)}} \cdot \frac{1}{x^2} - \frac{\sin{(x)}}{x} \cdot \frac{1}{x^2}  \end{align*}

    Go from there, using the well-known limit \displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} = 1 \end{align*}.
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    Re: Limit

    Quote Originally Posted by abualabed View Post
    hi , help please
    Limit-untitled2.gif
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  5. #5
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    Re: Limit

    hi
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    Re: Limit

    Quote Originally Posted by abualabed View Post
    hi
    Where on Earth did you get \displaystyle \begin{align*} 2 \times \frac{1}{4} \end{align*}?
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    Re: Limit

    palestine
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    Re: Limit

    Quote Originally Posted by abualabed View Post
    palestine
    a subtle error: limit for x --> 0 not (x/2) --> 0 not the same thing.
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  9. #9
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    Re: Limit

    if x-->0 then (x/2)-->0
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  10. #10
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    Re: Limit

    Quote Originally Posted by abualabed View Post
    if x-->0 then (x/2)-->0
    Which one converges faster to 0 ? this cannot be ignored in the serch for the limit of OP question.

    Limit-untitled2.gif
    Last edited by votan; October 1st 2013 at 05:14 AM.
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