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Math Help - Is this proof correct?

  1. #1
    Member
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    Feb 2010
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    146

    Is this proof correct?

    Hey guys I just need some one to check this proof. Its sort of my weakness so if someone could just double check it. and please forgive the minor symbol mistakes, I tried

    There are two parts to this problem but I got the first one with no problem but for the sake of the question I'm going to post everything.

    Let x_0 and L be real numbers, and let f be a real-valued function defined in a deleted neighborhood of x_0.

    (a) What is the logical negation of img.top {vertical-align:15%;} x_0" alt="lim_x-> x_0" />f(x)=L?

    The def is: For all \epsilon>0 there exists \delta>0 for all x such that 0<|x- x_0|< \delta \Rightarrow |f(x)-L|< \epsilon

    The negation is: There exists \epsilon>0 for all \delta>0 there exists x such that 0<|x- x_0< \delta and |f(x)-L|>= \epsilon

    b is the one I need checked:
    (b) Show that lim_x-> x_0f(x)=L if and only if for every sequence { x_n}n=1 to infinity with lim_x-> x_n= x_0 with x_n \neq x_0, it is true that lim_x->infinity f( x_n=L

    Here is my proof:
    Suppose img.top {vertical-align:15%;} x_0" alt="lim_x-> x_0" />f(x)=L. Let { x_n}n=1 to infinity be a sequence where x_n \neq x_0 but lim_n-> infinity x_n= x_0 and consider the sequence {f( x_n)}n=1 to infinity . choose \epsilon>0/ there is delta>0 such that if 0<|x- x_0|<delta then |f(x)-L|< \epsilon. Since {x_n} n=1 to infinity converges to x_0 there is N such that for n>=N, | x_n- x_0|<delta. Now for n>=N 0<| x_n- x_0|<delta. Hence, |f( x_n)-l|< \epsilon. Thus, {f(x_n)}n=1 to infinity converges to L.

    Suppose now that the second condition is satisfied and so all the sequences {f(x_n)}n=1 to infinity have a common limit called L. Suppose that L is NOT a limit of f at x_0. Thus, there is \epsilon>0 such that for every delta>0, 0<|x-x_0|<delta and such that |f(x)-l|>= \epsilon. In particular for each positive integer n, 0<| x_n- x_0)<1/n such that |f(x_n)-L|>= \epsilon. The sequence { x_n}n=1 to infinity converges to x_0 and is a sequence where x_n is NOT = x_0, hence {f(x_n)}n=1 to infinity converges to L which contradicts the fact that |f(x_n)-L|>= \epsilon>0 for all n. Thus L must be the limit of f at x_0.
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  2. #2
    Senior Member Tinyboss's Avatar
    Joined
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    Looks good, but work on your LaTeX! ;-)
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