Let f and g map a metric space into the reals, such that f and g have limits s and t as x approaches p. Does f×g have the limit s×t at p?
Since scaling is not a problem, multiply f or g by -1 as necessary, so that s and t are not negative.
If we stay close to p, f(x) can be bounded below s+1, which we will call b. If g aproaches 0, f×g is dominated by b×g, which is a scale multiple of g, and goes to 0. Thus if either f or g approaches 0 their product approaches 0, as expected.
Assume s and t are positive. Choose ε so that f(x) and g(x) are within ε of s and t respectively. How far away is f×g from s×t? The difference is no worse than (s+t)ε+ε2. We're only interested in ε < 1, so this is no worse than (s+t+1)ε. Restrict δ so that f and g are within ε/(s+t+1) of their limits, and the product is within ε of s×t.
Since complex multiplication is a linear combination of real multiplication, the limit of f(z)×g(z) is the complex product of the two limits.
If g approaches 0 the limit of f/g is not defined; but if g has a nonzero limit, the limit of the quotient is the quotient of the limits.
Since we can always multiply by f later, it is enough to consider 1/g. Let b be a positive number below the limit of g at p, and always restrict δ to a neighborhood where g(x) exceeds b. Further restrict δ so that g(x) is within ε of its limit t. Now subtract 1/t from 1/g(x), giving ε/(t×(t+ε)). This fraction is made even larger if we change the denominator to something smaller, namely b2. Thus the difference is a multiple of ε, and as ε shrinks, 1/g(x) approaches 1/t, which is 1/g(p).
Complex division is implemented by real operations, hence the limit of f(z)/g(z) is the complex quotient of the limits.
As a corollary, the product/quotient of continuous functions is continuous. But we can't make the same claim for uniformity. The identity function x is uniform; just set δ = ε. Yet 1/x is not uniform on (0,1), and x2 is not uniform over the entire x axis.
Recall that the proof of the product of limits uses the factor s+t+1. If f and g can be bounded, then s+t+1 can be bounded, and the product of uniform functions is uniform. Similarly, if g is bounded above 0 over its domain, and g is uniform, then 1/g is uniform.