To have a limit L as x approaches a, a function must be defined on both sides of a, and its value of f (x) must approach L as x approaches a from either side. Recall in supplemental notes 1 that we looked at a couple of limits approaching a from both sides. When the limits of both sides equaled each other, then we said that the limit exists. The ordinary limit that we have been studying is sometimes called two-sided limits. When we were investigating the limits as x approaches a from either the right- or left-hand side, these are what we call one-sided limits.

DEFINITION:

Let f (x) be defined on an interval (a, b) where a < b. If f (x) approaches arbitrarily close to L as x approaches a from within that interval, then we say that f has a right-hand limit L at a, and

Let f (x) be defined on an interval (c a) where c < a. If f (x) approaches arbitrarily close to M as x approaches a from within the interval (c, a), then we say that f has a left-hand limit M at a, and

FACT:

A function cannot have an ordinary limit at an endpoint of its domain, but it can have a one-sided limit.

If a function has an endpoint, and then that means any points beyond that point, the function is not defined. Therefore, a two-sided (or ordinary) limit cannot be used, but a one-sided can be used. A function that has an endpoint in its domain is a function that an even roots in it. Let us look at an example with a square root in it.

EXAMPLE 1:

SOLUTION:

two-sided limit would not make sense. Here is a theorem that relates one-sided and two-sided limits.

THEOREM:

A function f (x) has a limit as x approaches c if and only if it has left-hand and right-hand limits there, and these one-sided limits are equal.

EXAMPLE 2:

SOLUTION: When x ® -2 ^-, the values of x are smaller than -2, so the top will be positive and the bottom will be negative.

When x ® -2 ⁺, the values of x are larger than -2 (in the positive sense), so both top and bottom will be positive.

(If this was a two-sided limit, then the limit would not exist.)

EXAMPLE 3:

SOLUTION: Let us first rationalize the numerator, so I can factor out an h from the top and the bottom of the fraction.

EXAMPLE 4:

SOLUTION:

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

EXAMPLE 7:

SOLUTION: Since this is approaching 5 from the right-hand side, the top in absolute value will be positive (the square root of a square is the same as saying the absolute value) and the bottom will be negative.

Consider the function f (x) = 1/x (See figure 1) whose domain is (-¥ , 0) È (0, ¥ ). Let us determine the limit of the function as x approaches 0 from the right. (See table 1)

Therefore,

figure 1

x

1

0.5

0.1

0.01

0.001

0.0001

f (x)

1

2

10

100

1000

10000

As x gets closer to zero from the right-hand side, the f (x) values gets larger in the positive sense. Now let us determine the limit of the function as x approaches 0 from the left. (See table 2)

x

-1

-0.5

-0.1

-0.01

-0.001

-0.0001

f (x)

-1

-2

-10

-100

-1000

-10000

As x gets close to zero from the left-hand side, then f (x) values get larger in the negative sense. What is happening at x = 0? x = 0 is a vertical asymptote, and see figure 1 for a graph of this function.

EXAMPLE 8:

SOLUTION: Let us look at some values of x as x approaches 2 from the left.

x

1

1.5

1.75

1.9

f (x)

-3

-6

-12

-30

So as x gets close to 2 from the left, the f (x) values get large in a negative sense.

EXAMPLE 9:

SOLUTION: Here we are approaching -5 from the left-hand side, so again let us look at some values.

x

-6

-5.5

-5.1

-5.01

f (x)

9

16.5

76.5

751.5

EXAMPLE 10:

SOLUTION: First thing we should notice is that tangent is undefined at p /2. So let us look at the graph of this function on the interval (0, p /2).