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Math Help - Existence of limits!

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    Existence of limits!

    How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )
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    Quote Originally Posted by CrazyCat87 View Post
    How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )
    If x< 0, then x+ sgn(x)= x- 1. If x> 0, then x+ sgn(x)= x+1. For x very close to 0 and negative, x+ sgn(x) is very close to -1. For x very close to 0 and positive, x+ sgn(x) is very close to 1. For any putative limit, a, take \epsilon to be the smaller of |a-1|/2 and |a+1|/2. Depending on which you use, there will be either positive x or negative, arbitrarily close to x such that |x+ sign(x)- a| is larger than \epsilon.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )
    A slightly less messy approach would be to assume that \lim_{x\to0}\left\{x+\text{sgn}(x)\right\} exists and equals L. Then, L=L-0=\lim_{x\to0}\left\{x+\text{sgn}(x)\right\}-\lim_{x\to 0}x=\lim_{x\to0}\left\{x+\text{sgn}(x)-x\right\}=\lim_{x\to0}\text{sgn}(x). The combination of the limits was justified under the assumption that our limit exists. But the above then implies that \lim_{x\to0}\text{sgn}(x) exists, which by following HallsOfIvy's example is just ridiculous.
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