Thread: 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.

By definition of accumulation point there is a sequence of distinct points .
But by the given you also know that .
If it were the case that then there is an open set containing and no point of .
There is a contradiction there.

Originally Posted by noles2188

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 , 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