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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- Apr 7th 2009, 03:52 AMnoles2188Limits 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.

- Apr 7th 2009, 08:57 AMPlato
By definition of accumulation point there is a sequence of distinct points $\displaystyle \left( {d_n } \right) \subset D\;\& \;\left( {d_n } \right) \to c$.

But by the given you also know that $\displaystyle f\left( {d_n } \right) \to L$.

If it were the case that $\displaystyle L < a \vee b < L$ then there is an open set containing $\displaystyle L$ and no point of $\displaystyle [a,b]$.

There is a contradiction there. - Apr 9th 2009, 05:08 AMxalk
Suppose L<a ===> a-L>0............................................... ...........1

Now since $\displaystyle \lim_{x\rightarrow c}{f(x)} =L$, 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