The ratio f_n(x)/g_n(x) does NOT converge uniformly if at some x=x_0: g(x_0) = 0.

One of possible formulations to make the theorem CORRECT is the following:

"
There is a set X, there are sequences of functions f_n: X-->R and g_n: X-->R, both series converge uniformly, f_n-->f(x) and g_n-->g(x), and the limit functions have the following properties:
(a) both f and g are bounded, that is there are some F and G such that |f(x)| < F and |g(x)| < G for all x in X,
(b) g(x) has the property that there is some H>0 such that |g(x)| > H for all x in X.

Then f_n(x)/g_n(x) --> f(x)/g(x) uniformly.
"

This formulation can be modified.
Examples:
--- it is possible to replace the real numbers R by complex numbers C
--- it can be stated "if and only if".
--- condition (b) can be replaced by "the set X is closed and g(x) not = 0 in X"
+ there can be other modifications

... 3.1. This in turn is the same as the definition of uniform convergence of a sequence of ... The proof is given in part (aA) of Problem 11 on page 55, posting 240036 ...

... With metric defined by equation (0.1) it means simultaneous (uniform) convergence for each ... This provides an example of a proof regarding functions that satisfy ...

... are the best way to ensure uniform performance ... Poka-Yokes are mechanisms used to mistake-proof an entire ...Convergence means that fabricated parts converge into a ...

... That is, is there convergence at the top of the ... eliminated quotas on textile products; established more uniform standards for proof of dumping ...

... is true, one would expect a convergence of opinion ... inﬂuence is unlikely to have been uniform across all ... limited available data, show deﬁnitive proof of the ...

... Please see the attachment. Attachment #1 Proof: 1. I first show f is uniform continuous. ... Now let us back to the limit which means the uniformly convergence. ...

Series : Uniform Convergence and No Zeros. Fix R>0. Show that, if n is large enough, then P_n(z)=1+z+z^2/2!+z^3/3!+...z^n/n! has no zeros in {z:|z|<=R}. Proof: ...

... for all , then the series exhibits absolute convergence for each as well as uniform convergence in . ∞. ∑axn is convergent uniformly on [0,1]. Proof. ...