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Math Help - Limits of functions

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    Limits of functions

    Let f: D-->R and let c be an accumulation point of D. Suppose that a<=f(x)<=b for all x elements in D with x not equal to c, and that lim(x-->c)f(x)=L. Prove that a<=L<=b.
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    By definition of accumulation point there is a sequence of distinct points \left( {d_n } \right) \subset D\;\& \;\left( {d_n } \right) \to c.
    But by the given you also know that f\left( {d_n } \right) \to L.
    If it were the case that L < a \vee b < L then there is an open set containing L and no point of [a,b].
    There is a contradiction there.
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    Quote Originally Posted by noles2188 View Post
    Let f: D-->R and let c be an accumulation point of D. Suppose that a<=f(x)<=b for all x elements in D with x not equal to c, and that lim(x-->c)f(x)=L. Prove that a<=L<=b.
    Suppose L<a ===> a-L>0............................................... ...........1

    Now since \lim_{x\rightarrow c}{f(x)} =L, we HAVE that for every +ve NO (and thus for a-L) ,THERE exists a δ>0 and such that:

    if xεD and 0<|x-c|<δ,then |f(x)-L|<a-L.

    But : |f(x)-L|<a-L <====> L-a<f(x)-L<a-L ====> f(x)<a ,a contradiction.

    In the same way we work for b
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