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Math Help - "Squeeze/Sandwich Theorem" I'm lost

  1. #1
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    "Squeeze/Sandwich Theorem" I'm lost

    Hi, is it just me or do text books don't explain the Squeeze/Sandwich theorem properly? I'm always unsure of the domain required. Some help would be very much appreciated =]
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  2. #2
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    The domain isn't really much of a problem, so long as the two bounding functions converge there. The sandwich principle says

    If a function f(z) is bound between two convergent functions g(z),h(z) that converge to l_g,l_h respectively then l_f is somewhere between l_g and l_h (providing that f(z) is a convergent function). Hence, if l_g=l_h then l_f=l_g=l_h

    Please observe that, if a function is bound between two functions that do not converge to the same limit, this does not mean that the first function converges.
    Take, for example f(x)=\sin(x) and g(x)=\frac{3x^2+3x}{(2x+1)(x+4)},h(x)=\frac{(1-x)(2+2x^2)}{x^3}. Clearly \lim_{n\rightarrow\infty}g(x)=3/2, \lim_{n\rightarrow\infty}h(x)=-2 and l_h\leq-1\leq\sin(x)\leq 1 \leq l_g.
    Just because f(x) is bound between two convergent series does not mean that it converges

    Paul
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  3. #3
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    hrmm...i think i wasn't clearly pointing out the problem . Let's say we try to prove:

    lim x-> 0 x.cos(1/x) = 0

    squeeze theorem states that: f(x) <= g(x) <= h(x)

    and
    limf(x) = lim(hx) = L
    then limg(x) = L

    i don't know where to begin:

    i know you must find a domain for which x.cos(1/x) lies in and then take the limits of the left bound and right bound, which would equal to the limit of the function being squeezed.

    But, how to find the domains? Thanks for the reply earlier Paul.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by tasukete View Post
    lim x-> 0 x.cos(1/x) = 0

    squeeze theorem states that: f(x) <= g(x) <= h(x)

    and
    limf(x) = lim(hx) = L
    then limg(x) = L
    I'll try to help you out quickly.

    Let's first work with \cos \frac{1}{x}

    The cos graph will be oscillating between [-1 ; 1]

    Therefore:

    -1 \leq \cos \frac{1}{x} \leq 1

    Multiply by x through the inequality.

    -x \leq x \cos \frac{1}{x} \leq x

    If x approaches zero, we obtain the following:

    0 \leq x \cos \frac{1}{x} \leq 0

    And if the left and right is zero, then the middle must also be zero... (That's logical! )
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  5. #5
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    so for these kind of problems you've got a have a mental picture of the graph? i've done a couple of problems and starting to get the feel of it now. thnx Pual =]
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by tasukete View Post
    so for these kind of problems you've got a have a mental picture of the graph?
    Not really. You know very well that sin and cos oscillates between [-1 ; 1]

    Quote Originally Posted by tasukete View Post
    i've done a couple of problems and starting to get the feel of it now.
    Practise makes perfect.

    Quote Originally Posted by tasukete View Post
    thnx Pual =]
    Not Paul!
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  7. #7
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    Wink

    Sorry. LOL pls ignore that...
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