Bounded function implies limit is bounded

miren324
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Here's the problem:
Let f:D-R and c in R be and a accumulation point of D, which is a subset of R. Suppose that a<=f(x)<=b for all x in D, x not equal to c, and suppose that limx\rightarrowc f(x) = L. Prove that a<=L<=b.

I'm having trouble here. I've tried to prove by contradiction, by assuming that L<a, then after a contradiciton, assuming L>b. This led through a lot of junk and I ended up back where I started. I am unsure how to construct a proof using the analogous relationship of sequences and functions.

Any help would be greatly appreciated.
 
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What is your intuition? If you were going to explain informally the reason, what would you say?
 
It just makes sense that it would be true. I know that if x does not equal c, the limit as x approaches c of f(x) is L, and L is "near" f(c-.00001) and f(c+.00001) and f(c-.000000001), etc. But I only know that from experience with linear equations. I'm not sure how to translate this into a formal proof for all functions, domains, and limits.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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