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Math Help - limit of function of a sequence equals the function of the limit of that sequence?

  1. #1
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    limit of function of a sequence equals the function of the limit of that sequence?

    ok I'm doing my homework when i face this problem:

    suppose f is continuous at x = c in [a,b]. this implies that for any sequence {xn} in [a,b] and {xn} converges to c, we have the sequence {f(xn)} converges to f(c). In other words: lim f(xn) = f(lim (xn)) with n approaches to infinity. True or False?

    I think it is true so maybe I can use definition to prove it? But I'm not sure.
    f continuous at c so by definition, for any e>0 there exists alpha > 0 such that |x-c| < alpha >> |f(x)-f(c)|< e

    and {xn} converges to c so also by definition for every e > 0, there is a n* such that for all n >= n* >> | xn - c| < e

    how can i go from here to get to:

    |f(xn) - f(c)| < e ?

    Thanks very much in advance!
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  2. #2
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    Just let \alpha  = \varepsilon~.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Just let \alpha  = \varepsilon~.
    so it is just like this?

    alpha = e

    so |xn - c| < alpha so for all e > 0 there exists n* in N such that for all n >= n*
    |f(xn) - f(c) | < e

    >> f(xn) converges to f(c)
    >> lim (f(xn) = f(c) = f(lim (xn)) as n approaches to infinity.

    Thanks very much for your reply.
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