Now we can extend concept of limit.
Definition. We write that limx→a−f(x)=L and say "the limit of f(x), as x approaches a from the left, equals L" if for any ϵ>0 there exists δ>0 such that ∣f(x)−L∣<ϵ when ∣x−a∣<δ and x>a.
In other words we are interested in behavior of function near a and only to the left from a.
Definition. We write that limx→a+f(x)=L and say "the limit of f(x), as x approaches a from the right, equals L" if for any ϵ>0 there exists δ>0 such that ∣f(x)−L∣<ϵ when ∣x−a∣<δ and x<a.
In other words we are interested in behavior of function near a and only to the right from a.
One-sided limits often arise for piecewise defined functions.
Example 1. Find limx→1+f(x), limx→1−f(x), limx→1f(x) where f(x)={2ifx≥11ifx<1.
As x approaches 1 from the left f(x) approaches 1, thus limx→1−f(x)=1.
As x approaches 1 from the right f(x) approaches 2, thus limx→1+f(x)=2.
So, there is no single number that f(x) approaches as x approaches 1, thus limx→1f(x) doesn't exist.
Fact. limx→af(x)=L if and only if limx→a−f(x)=limx→a+f(x)=L.
This fact is quite useful because often it is easier to find one-sided limits and to check whether they are equal to make conclusion about limit.
Example 2. The graph of the function f(x) is shown. Find (if they exist) values of the following limits:

- limx→1−f(x)
- limx→1+f(x)
- limx→1f(x)
- limx→4−f(x)
- limx→4+f(x)
- limx→4f(x)
From graph it is seen that f(x) approaches 2 as x approaches 1 from the left and f(x) approaches 3 when x approaches 1 from the right.
Therefore, limx→1−f(x)=2 and limx→1+f(x)=3.
Since limx→1−f(x)=limx→1+f(x) then limx→1f(x) doesn't exist.
Also from graph it can be seen that f(x) approaches 1 as x approaches 4 from the left and f(x) approaches 1 when x approaches 4 from the right. Therefore, limx→4−f(x)=1 and limx→4+f(x)=1.
Since limx→4−f(x)=limx→4+f(x)=1 then limx→4f(x)=1.
Despite this fact notice that f(4)=1.
Example 3. Find limx→0∣x∣.
Recall that ∣x∣={xifx≥0−xifx<0.
Therefore, for x>0 ∣x∣=x, thus limx→0+∣x∣=limx→0+x=0.
Since for x<0 ∣x∣=−x, then limx→0−∣x∣=limx→0−−x=0.
We see that limx→0+∣x∣=limx→0−∣x∣=0. That's why limx→0∣x∣=0.
Example 4. Find limx→0x∣x∣.
For x>0 ∣x∣=x, thus limx→0+x∣x∣=limx→0+xx=limx→0+1=1.
For x<0 ∣x∣=−x, thus limx→0−x∣x∣=limx→0−x−x=limx→0−−1=−1.
We see that limx→0+x∣x∣=limx→0−x∣x∣. That's why limx→0x∣x∣ doesn't exist.