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Math 231E, Lecture 5.
One-sided limits, and Limit Laws
1 One-sided limits
We can also weaken the notion of limit, and only require that the limit work “on one side” of the point in question. This is why we define limits “from the left” and “from the right”. We say that lim x→a+ f (x) = L or lim x↘a
f (x) = L ,
if,
for all > 0 , there is a δ > 0 such that if a < x < a + δ, then |f (x) − L| < .
This is called the “right-hand limit”. We can also consider the left-hand limit: We say that lim x→a− f (x) = L or lim x↗a f (x) = L ,
if,
for all > 0 , there is a δ > 0 such that if a − δ < x < a, then |f (x) − L| < .
Consider the function
f (x) =
0 , x < 0 , 1 , 0 ≤ x < 1 , 2 x + 1, 1 ≤ x,
which has a graph that looks like:
x
f (x)
Let us now choose a = 1. Our intuition tells us that
lim x→ 1 − f (x) = 1, lim x→1+ f (x) = 3.
Let us prove the latter. We want to show that for any > 0 , we can find a δ so that if 1 < x < x + δ, then |f (x) − 3 | < . Noting f (x) = 2x + 1, we have f (x) − 3 = 2x − 2 if x > 1. So we have
|f (x) − 3 | < | 2 x − 2 | < 2 − < 2 x < 2 + 1 −
< x < 1 +
Thus we need to satisfy
x > 1 , and x < 1 +
and it is clear that if we choose δ = / 2 , then we have satisfied the right-hand limit condition. Here is the following important statement:
lim x→a f (x) = L
if and only if
lim x→a− f (x) = L, and lim x→a+ f (x) = L.
In short, if the left-hand limit exists, and the right-hand limit exists, and they are equal, then the limit exists.
2 Infinite limits
Finally, we want to make sense of the notation
lim x→a f (x) = ∞ ,
and we define it as such:
for all M > 0 , there is a δ > 0 such that f (x) > M whenever 0 < |x − a| < δ
Example 1. We will prove that
lim x→ 0
x^2
Choose any M > 0. We need to show find a δ such that |x| < δ implies
1 x^2
> M,
lim x→a (f (x)g(x)) = lim x→a f (x) · lim x→a g(x)
lim x→a
f (x) g(x)
limx→a f (x) limx→a g(x)
, but only if lim x→a g(x) 6 = 0!
lim x→a (cf (x)) = c lim x→a f (x)
lim x→a (f (x))n^ =
lim x→a f (x)
)n
lim x→a c = c
lim x→a x = a
lim x→a xn^ = an
lim x→a
√ nx = √na
Example 2.
lim x→ 3 3 x^2 − 4 x + 2 = lim x→ 3 3 x^2 − lim x→ 3 4 x + lim x→ 3
= 3 lim x→ 3 x^2 − 4 lim x→ 3 x + lim x→ 3
(In short, just plug in!)
Example 3. Let us consider some rational functions:
lim x→ 1
x^2 − 2 x + 7 x + 3
limx→ 1 x^2 − 2 x + 7 limx→ 1 x + 3
lim x→ 1
x^2 − 2 x + 1 x + 3
limx→ 1 x^2 − 2 x + 7 limx→ 1 x + 3
lim x→ 1
x^2 − 2 x + 7 x − 1
limx→ 1 x^2 − 2 x + 7 limx→ 1 x − 1
, Law #4 fail
lim x→ 1
x^2 − 2 x + 1 x − 1
limx→ 1 x^2 − 2 x + 1 limx→ 1 x − 1
, Law #4 fail
4 Piecewise limits
Example 4. What about f (x) = |x| near 0? None of the Limit Laws work. We have to break the function up into pieces. However, notice that
|x| =
x, x > 0 , −x, x < 0.
So, we can compute lim x→0+ |x| = lim x→0+ x = 0,
and lim x→ 0 −
|x| = lim x→ 0 −
(−x) = 0,
and since both left- and right-hand limits exist, and are the same, we can say that
lim x→ 0 |x| = 0.
Question : What happens when we try f (x) =
|x| x
