What is the Multiplication Rule for Limits? - Definition & Concept

Instructor: Glenda Boozer
If we know the limit of two functions at the same particular value, what does that tell us about the limit at that same value of the product of the two functions? Let's look at the Multiplication Rule for limits.


The multiplication rule for limits says that the product of the limits is the same as the limit of the product of two functions. That is, if the limit exists and is finite (not infinite) as x approaches a for f(x) and for g(x), then the limit as x approaches a for fg(x) is the product of the limits for f and g.

In case you need some background on limits, consider the following:

Most of the time, a function f(x) has a value for every possible value of x, but not always. For example, f(x) = (x - 4)(x - 6)/2(x - 6) is undefined at the value x = 6 because dividing by 2(6 - 6) = 0 is just not feasible. Even then, however, we can look at what the function is like when we get closer and closer to the limit: in this case, the value x = 6. As we can see below, the closer x gets to 6, the closer y gets to 1, so we say that the limit of f(x) as x approaches 6 is 1. We can define a limit if we can get as close as we want to a particular value of x (call it x sub 0) and if it consistently gets closer and closer to one particular value of y when we do. If we can't get close to the x value with a definition of a function, or if the values of y jump around when we get closer and closer to x sub 0, or if the values of y go up or down without ever settling near a number, we cannot define a limit.

limit diagram

Multiplication Property at Work

Now, suppose we have two functions, and both have limits at a certain value. For example, suppose f(x) = (x^2 - 1)/(x + 1) and g(x) = x + 3. Each of them has a limit at x = -1: the limit of f(x) as x approaches -1 is -2, and the limit of g(x) as x approaches -1 is 2. Let's look at the graphs:

f(x) with limit

Note that f(x) can be simplified to (x + 1)(x - 1)/(x + 1) = x - 1, but we have to leave the function undefined at x = -1.

g(x) with limit

So far, so good. Notice that g(x) has a defined value at x = -1 and f(x) does not. The limits still work out just fine.

Now, suppose we look at fg(x). We can multiply most functions by multiplying the expressions that define them, so in our example,

fg(x) = (x + 3) (x^2 - 1)/(x + 1)= (x^3 - x)/(x + 1) + (3x^2 - 3)/(x + 1).

Note that this will simplify to x(x - 1) + 3(x - 1) = x^2 - x + 3x - 3 = x^2 + 2x - 3, but x = -1 will still be undefined.

Let's graph the new function and see what happens to the limit as x approaches -1:

fg(x) with limit

This is part of a parabola, as we would expect from a quadratic function. Of course, fg(x) is not defined for x = -1. The limit as x approaches -1 is -4.

So the limit as x approaches -1 for f(x) is -2, the limit as x approaches -1 for g(x) is 2, and the limit as x approaches -1 of fg(x) is -4, which is (-2)(2).

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