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Math Help - Trig limit

  1. #1
    Member Marconis's Avatar
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    Trig limit

    The question is:

    lim t-->0 (sin^2 3x / x^2)

    How do you break it up? The sin^2 is throwing me off. Do you do this?

    3 (sinx . sinx / x . x). Then get the answer is 3?

    I understand the sinx/x=1 rule. PLEASE guide me! I am good at calculating limits, but trig always makes me depressed
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  2. #2
    MHF Contributor

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    Take note: \displaystyle\lim _{x \to 0} \frac{{\sin (3x)}}<br />
{x} = 3
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  3. #3
    Member Marconis's Avatar
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    Yes.
    So am I right?
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  4. #4
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    Quote Originally Posted by Marconis View Post
    The question is:

    lim t-->0 (sin^2 3x / x^2) should that be x-->0 ?

    How do you break it up? The sin^2 is throwing me off. Do you do this?

    3 (sinx . sinx / x . x). Then get the answer is 3?

    I understand the sinx/x=1 rule. PLEASE guide me! I am good at calculating limits, but trig always makes me depressed
    you could write also

    \displaystyle\lim_{x \to 0}\frac{Sin^2y}{\left(\frac{y}{3}\right)^2}=\lim_{  y \to 0}3^2\frac{Siny}{y}\ \frac{Siny}{y}
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  5. #5
    Flow Master
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    Quote Originally Posted by Marconis View Post
    Yes.
    So am I right?
    No. From what is said in post #2, it should be clear that the answer will be 3^2.
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  6. #6
    Member Marconis's Avatar
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    Ah, alright. Thanks guys!
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