Uniform convergence of a quotient

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Homework Statement


Let f,g be continuous on a closed bounded interval [a,b] with |g(x)| > 0 for all x in [a,b]. Suppose that f_n \to f and g_n \to g uniformly on [a,b]. Prove that \frac{1}{g_n} is defined for large n and \frac{f_n}{g_n} \to \frac{f}{g} uniformly on [a,b]. Show that this is not true if [a,b] is replaced with (a,b).


Homework Equations





The Attempt at a Solution


The fact that g_n \to g uniformly coupled with |g(x)| > 0 is enough for g_n \neq 0 for large enough n, which means that \frac{1}{g_n} is defined. I'm stuck on the other part. The fact that it is seemingly not true for an open interval domain suggests that I need that the limit functions are bounded, but I've not read anything that says a continuous limit implies continuous sequence elements, even under uniform convergence. The converse is of course true, but I'm not sure about this direction of implication. I'm really just stuck on even where to begin trying to prove that |\frac{f_n}{g_n} - \frac{f}{g}| < \epsilon.

Any help whatsoever is greatly appreciated.
 
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You are going to need more than ##g_n(x)\ne 0## for n large. Can you show there is ##d>0## such that ##g_n(x)\ge d > 0## for large n? You need that to show ##1/g_n(x)## is bounded. That should help.
 
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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