Question: Suppose that $(f_{k})$ converges uniformly to $f$ on a compact subset $k$ of $\mathbb{R}^n$ and that $g_{k}$ converges uniformly on $K$ to a continuous function $g$ such that $g(x) \neq 0$ for all $ x\in K$. Prove that $\frac{f_{k}(x)}{g_{k}(x)}$ is everywhere defined for large $k$ and that this quotient converges uniformly to $\frac{f(x)}{g(x}$ on $K$.
I am having two issues with this question:
1) I'm not sure what "proving the function is everywhere defined" means in terms of satisfying a definition.
2) More importantly to me I'm butchering how to obtain the necessary bounds for the quotient. After a bit of algebra I will arrive here:
$$\|\frac{f_{k}(x)}{g_{k}(x)} - \frac{f(x)}{g(x)}\| \Rightarrow \frac{\|g(x)\| \|f_{k}(x) - f(x) \| + \|f(x)\| \|g_{k}(x) - g(x)\|}{\|g_{k}(x)\| \|g(x)\|}$$
This is where my problem lies. How do I obtain a bound for $\|g_{k}(x)\|$? In theory it should be very similar to the case proven for sequences of numbers, but I am getting utterly confused on how to treat the sequence of functions.
So if this were a sequence of numbers instead of functions, by the fact that the sequences converge it implies they are bounded, but I don't think that idea translates directly and there is a nuance in selecting my $N > 0$ to satisfy the convergence.
EDIT: Also is $\|g(x)\|$ AND $\|f(x)\|$ just fixed numbers or are they things that also need to get bounded?
$\|g(x)\|$ attains its minimum value (by compactness of $K$ and continuity). Let $m$ be this minimum value. Choose $N$ such that $\|g_n(x)-g(x)\| <m/2$ for $n \geq N$. Then $|g_n(x)| \geq |g(x)| -|g_n(x)-g(x)|>\frac m 2$ so $n \geq N$ which implies $g_n(x) \neq 0$ for every $x$. This is what is meant by saying that $\frac {f_n} {g_n}$ is everywhere defined as long as $n \geq N$. We have already seen that $|g_n(x)|>\frac m 2$ for all $x$ if $n \geq N$ and this should help you to complete the argument.