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Math Help - Limits

  1. #1
    pkr
    pkr is offline
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    Limits

    find the limit as x-> inf for f(x)= cos(x)/sqrt(x)

    so, I can tell it's going to be 0, but as a proof i'm not sure what to say

    I started by changing it to sqrt(x)*cos(x)/x <= sqrt(x)/(x) = 1/sqrt(x)

    Which obviously goes to zerp as x goes to infinity...but what about a lower bound?


    Another problem is lim x -> 0 for f(x) = sin^2(sin(x))

    All i've really got is that it is bounded below by 0 and above by 1, not really sure if a epsilon delta proof would be better...
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by pkr View Post
    find the limit as x-> inf for f(x)= cos(x)/sqrt(x)

    so, I can tell it's going to be 0, but as a proof i'm not sure what to say

    I started by changing it to sqrt(x)*cos(x)/x <= sqrt(x)/(x) = 1/sqrt(x)

    Which obviously goes to zerp as x goes to infinity...but what about a lower bound?
    \lim_{x\to\infty}\frac{-1}{\sqrt{x}}\leq\lim_{x\to\infty}\frac{\cos(x)}{\s  qrt{x}} \leq \lim_{x\to\infty}\frac{1}{\sqrt{x}}

    Thus 0\leq\lim_{x\to\infty}\frac{\cos(x)}{\sqrt{x}}\leq  0, so the limit equals zero by the squeeze theorem.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by pkr View Post
    Another problem is lim x -> 0 for f(x) = sin^2(sin(x))

    All i've really got is that it is bounded below by 0 and above by 1, not really sure if a epsilon delta proof would be better...
    f(x)=\sin^2(\sin(x))

    This function is continuous, as it is the composition of two continuous functions, and thus \lim_{x\to a}f(x) = f(a)

    Using a=0, we have \lim_{x\to0}\sin^2(\sin(x)) = \sin^2(\sin(0)) = 0
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