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Math Help - The Squeeze Theroem

  1. #1
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    The Squeeze Theroem

    Hi, i got a question about the squeeze theorem.
    Suppose |f(x)|<=g(x) for all x. What can you conclude about lim(x-->a) f(x) if lim (x-->a) g(x) = 0? What if lim (x-->a) g(x)=3?


    What i think i should do is to set it up this way:
    -g(x)<=f(x)<g(x) to use the squeeze theorem. and i guess if g(x)=0, then -g(x)=0. But im not sure, and im not sure what happens if lim(x-->a) g(x)=3 with f(x)...

    Thanks for help!
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  2. #2
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    Quote Originally Posted by fredrikhoeg View Post
    Hi, i got a question about the squeeze theorem.
    Suppose |f(x)|<=g(x) for all x. What can you conclude about lim(x-->a) f(x) if lim (x-->a) g(x) = 0? What if lim (x-->a) g(x)=3?


    Since g(x)\xrightarrow [x\to a] {}0\Longrightarrow |g(x)|\xrightarrow [x\to a]{}0 , we'd get that

    |f(x)|\leq g(x)\Longleftrightarrow -g(x)\leq f(x)\leq g(x) , and now apply the squeeze theorem to get the result for \lim\limits_{x\to a}f(x) .

    Can you see what doesn't work if the limits is 3?

    Tonio



    What i think i should do is to set it up this way:
    -g(x)<=f(x)<g(x) to use the squeeze theorem. and i guess if g(x)=0, then -g(x)=0. But im not sure, and im not sure what happens if lim(x-->a) g(x)=3 with f(x)...

    Thanks for help!
    .
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  3. #3
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    Thanks, so I understand that since lim(x-->a) g(x)=0, the lim(x-->a) f(x) = 0, since lim(x-->a) -g(x) is also 0.

    hmm, what doesnt work when lim(x--a) g(x)=3? maybe that lim(x-->a) -g(x) isnt the same anymore?
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  4. #4
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    Yes, \lim_{x\to 3} g(x) \ne \lim_{x\to 3} -g(x) so f(x) isn't being "squeezed" at x=3.
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