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Extra Credit Problem from a past test

  1. Nov 23, 2008 #1
    Messed this up.
     
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 23, 2008 #2

    Dick

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    I don't know. I think you'd better post the full question. I can't make much sense out of that. I can't even figure out what the actual question is.
     
  4. Nov 24, 2008 #3
    If the limit exists it must sure be L. Just define F by F(Xn)=F(Yn)=L and F(x)=0 else. Then F certainly does not have a limit as x-->c.
     
  5. Nov 24, 2008 #4

    HallsofIvy

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    I think what you are asking is this: if [itex]\lim_{n\rightarrow\infty}x_n= c[/itex] and [/itex]\lim_{y_n\rightarrow 0} y_n= c[/itex], [itex]\lim_{n\rightarrow\infty}f(x_n)= L[/itex], and [itex]\lim_{n\rightarrow\infty}f(y_n)= L[/itex], is it necessarily true that [itex]\lim_{x\rightarrow c}f(x)= L[/itex]?

    The answer is no. What is true is that [itex]\lim_{x\rightarrow c} f(x)= L[/itex] if and only if [itex]\lim_{n\rightarrow\infty} f(x_n}= L[/itex] for every sequence [itex]{x_n}[/itex] that converges to c. Just two sequences isn't enough.

    For example, suppose f(x)= 0 if x is rational, 1 if x is irrational. Let [itex]{x_n}[/itex] and [itex]{y_n}[/itex] be two different sequences of rational numbers converging to 1. Then [itex]\lim_{n\rightarrow\infty} f(x_n)= \lim_{n\rightarrow\infty} g(x_n)= 0 but [itex]\lim_{x\rightarrow 1} f(x)[/itex] does not exist.

    Of course, if we know that [/itex]\lim_{x\rightarrow c} f(x)[/itex] exists then one such sequence is sufficient to tell us what the limit is.
     
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