Limits and Continuity, Multiplying and Dividing Limits

Multiplying and Dividing Limits

In this page, the range consists of real or complex numbers. Thus points in the range can be multiplied together, and it is meaningful to talk about f×g or f÷g.

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.