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Math Help - Limits

  1. #1
    Senior Member Twig's Avatar
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    Limits

    Hi!

    I need some help to get me started on these  \lim_{x\rightarrow 0} expressions.

    I just want to learn the mindset behind solving these, so I can try to solve a few on my own.

     \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}

    I want to decide the limit when  x\rightarrow \ 0

    Thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Twig View Post
    Hi!

    I need some help to get me started on these  \lim_{x\rightarrow 0} expressions.

    I just want to learn the mindset behind solving these, so I can try to solve a few on my own.

     \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}

    I want to decide the limit when  x\rightarrow \ 0

    Thank you
    Just remember that \lim_{u\to{0}}\frac{\sin(u)}{u}=1

    That should pretty much lead you to the answer you are looking for.

    --Chris
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  3. #3
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    hi

    Hi!

    Uhm.. letīs see here.

     \lim_{x\rightarrow 0} \ \frac{sin 3x}{3x}

    If I let  3x = v

    For every positive number  \epsilon there is a positive number
     \delta so that  0 < |\alpha - 0| < \delta so that  |sin (\alpha) - 1| < \epsilon

    Meaning in this case that for every angle v,  sin (v) = \ \mbox{some value}
    ,there is always an angle closer to 0 that makes  sin (\alpha) =  \ \mbox{lower value}

    Is this correct?

    I donīt really get how to proceed...

     sin (v) = 1 - \epsilon
    then I know from the definitions above that  0 < sin (\alpha) < 1 - \epsilon , since  \alpha \ \mbox{is an angle closer to 0, thus making the sine of the angle closer to 0}

    From here I donīt really know what to do...

    Thanks
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