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Uniform limit of continuous functions

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Problem is in the attached file.
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Please see the attachment file.

We need to show that for every epsilon>0 there's delta>0 such that for all x with
|x-c|<delta we have
|f(x)-f(c)|<epsilon

To get an idea for the proof, write
|f(x)-f(c)| = |fn(x)-fn(x)+fn(c)-fn(c)+f(x)-f(c)| <= |fn(x) - f(x)|+|fn(x)-fn(c)|+|fn(c)-f(c)|
Now, if we can estimate each term by an arbitrary ...

Solution Summary

A rigorous proof that a uniform limit of a sequence of continuous functions is continuous, is given.

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