Calculus- Early Transcendentals, 2021a

3.2. Precise Definition of a Limit 79
Proof. We must use the Precise Definition of a Limit to prove the Product Law for Limits. So given any ε
we need to find a δ so that 0 < |x − a| < δ implies | f (x)g(x) − LM| < ε. Whatdowehavetoworkwith?
We know that we can make f (x) close to L and g(x) close to M, and we have to somehow connect these
facts to make f (x)g(x) close to LM.
We use, as is often the case, a little algebraic trick:
| f (x)g(x) − LM| = | f (x)g(x) − f (x)M + f (x)M − LM|
= | f (x)(g(x) − M)+(f (x) − L)M|
≤|f (x)(g(x) − M)| + |( f (x) − L)M|
= | f (x)||g(x) − M| + | f (x) − L||M|.
This is all straightforward except perhaps for the “≤”. That is an example of the triangle inequality,
which says that if a and b are any real numbers then |a + b| ≤|a| + |b|. If you look at a few examples,
using positive and negative numbers in various combinations for a and b, you should quickly understand
why this is true. We will not prove it formally.
Suppose ε > 0. Since lim f (x)=L, there is a value δ 1 such that 0 < |x − a| < δ 1 implies | f (x) − L| <
x→a
ε
2(1 + |M|) . This means that 0 < |x − a| < δ 1 implies | f (x) − L||M| < | f (x) − L|(1 + |M|) < ε/2.
Now we focus our attention on the other term in the inequality, | f (x)||g(x)−M|. We can make |g(x)−
M| smaller than any fixed number by making x close enough to a; unfortunately, ε/(2 f (x)) is not a fixed
number, since x is a variable. Here we need another little trick, just like the one we used in analyzing
x 2 . We can find a δ 2 so that |x − a| < δ 2 implies that | f (x) − L| < 1, meaning that L − 1 < f (x) < L + 1.
This means that | f (x)| < N, whereN is either |L − 1| or |L + 1|, depending on whether L is negative or
positive. The important point is that N doesn’t depend on x. Finally, we know that there is a δ 3 so that
0 < |x − a| < δ 3 implies |g(x) − M| < ε/(2N). Letδ be the smallest of δ 1 , δ 2 ,andδ 3 .Then|x − a| < δ
implies that | f (x) − L| < ε/(2(1 + |M|)), | f (x)| < N, and|g(x) − M| < ε/(2N). Then
| f (x)g(x) − LM|≤|f (x)||g(x) − M| + | f (x) − L||M|
< N ε
2N + ε
(1 + |M|)
2(1 + |M|)
= ε 2 + ε 2 = ε.
This is just what we needed, so by the official definition, lim f (x)g(x)=LM.
x→a
The concept of a one-sided limit canalsobemadeprecise.
♣
Definition 3.7: One-sided Limit
Suppose that f (x) is a function. We say that lim f (x) =L if for every ε > 0 there is a δ > 0 so
x→a− that whenever 0 < a − x < δ, | f (x) − L| < ε. We say that lim f (x)=L if for every ε > 0 there is a
x→a +
δ > 0 so that whenever 0 < x − a < δ, | f (x) − L| < ε.