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Math Help - Limits (Squeeze)

  1. #1
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    Limits (Squeeze)

    I don't seem to understand the following problem:



    From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

    So, in other words, I'm completely lost.

    NEVERMIND I FIGURED IT OUT!!! So easy .
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  2. #2
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    Quote Originally Posted by BeSweeet View Post
    I don't seem to understand the following problem:



    From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

    So, in other words, I'm completely lost.
    Just apply the squeeze theorem!
    If \lim_{x \to a} \left(b - |x-a|\right) = Number = \lim_{x \to a} \left(b + |x-a|\right)
    Then, the desired limit = this number.
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  3. #3
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    Quote Originally Posted by BeSweeet View Post
    I don't seem to understand the following problem:



    From all the problems that I've analyzed and completed, they all involve numbers, not just letters/variables.

    So, in other words, I'm completely lost.
    \lim_{x->a}(b-\mid x-a \mid)=b-\mid a-a \mid =b

    Also

    \lim_{x->a}(b+\mid x-a \mid)=b

    So this means that b\leq \lim_{x->a}f(x)\leq b

    Therefore \lim_{x->a}f(x)=b by the squeeze theorem. Does that make sense?

    I have this same problem in the ninth edition of Calculus by Larson and Edwards. Are you working out of that same book? lol. I just did this problem for my homework last week, maybe we're in the same class.
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  4. #4
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    Given that inequality, we'll need to study the one sided limits.

    If x>a, then |x-a|=x-a, thus b-(x-a)\le f(x)\le b+(x-a) so by takin' the limit on both sides we arrive that \lim_{x\to a^+}f(x)=b; on the other side, if x<a, then |x-a|=a-x, so b-(a-x)\le f(x)\le b+(a-x) and then \lim_{x\to a^-}f(x)=b, so both sided limits are equal and then f(x)\to b as x\to a.
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  5. #5
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    Read the edit . I guess I was a bit too lazy to even look over the problem a few times.
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