Consider the function

and its graph (See fig 1). Let us look at the limit of the function as x goes to ± ¥ . To do this, I will construct a table of values, then I will make my conclusions.

figure 1

x

1

8

98

998

-12

-102

-1002

f (x)

1/3

0.1

0.01

0.001

-0.1

-0.01

-0.001

The above is an example of a limit of a function as x approaches ± ¥ . The properties of limits that we learned in supplemental notes 2 still apply in this case. So let us work some examples.

EXAMPLE 1:

SOLUTION:

EXAMPLE 2:

SOLUTION:

EXAMPLE 3:

SOLUTION: To find this limit, I will have to apply the sandwich theorem. First of all, we know that -1 £ sin x £ 1.

Therefore, by the sandwich theorem,

EXAMPLE 4:

SOLUTION:

FACT:

¥ / ¥ is an indeterminate form.

FACT:

So how do we find this limit? First of all, note that the degree of the top and bottom are the same. So we will divide both top and bottom by the highest degree term which is x ².

EXAMPLE 5:

SOLUTION: First of all, note that the degree of the numerator is smaller than the degree of the denominator. So I will divide both top and bottom by the highest degree term of the denominator.

EXAMPLE 6:

SOLUTION: Notice that in this case, the degree of the numerator is higher that the degree in the denominator. In this case I am going to divide both top and bottom by the highest degree term of the denominator.

These last three examples should reveal a pattern to you.

Here is the pattern.

1.

If the numerator and the denominator have the same degree, then the limit is the ratio of the polynomial's leading coefficients.

2.

If the degree of the numerator is smaller than the degree of the denominator, then the limit will be zero.

3.

If the degree of the numerator is greater than the degree of the denominator, then the limit will be ± ¥ .

EXAMPLE 7:

SOLUTION: The degree of the numerator and the denominator are the same, so this limit fits case 1.

EXAMPLE 8:

SOLUTION: We have the following degrees in this problem: -1, 0, 1/2, 1. The highest degree term is in the denominator, so I am going to divide everything by x.

When a limit goes to ± ¥ as x ® a, then the function can be doing one thing- it is approaching a vertical asymptote. When a limit approaches a fixed value as x ® ± ¥ , then the function is approaching a horizontal asymptote.

Now let us make some connections about domains, continuity, vertical asymptotes, and limits that approach ± ¥ . First of all, any point not contained in the domain will be a point of discontinuity. If you recall, there are two basic categories of discontinuities- removable points of discontinuity and vertical asymptotes. Next you will discover, that as x approaches the vertical asymptote from the left or the right, the limit will go to either ± ¥ .

What happens to a function as its limit approaches ± ¥ as x goes off to ± ¥ ? This is the third case in which the degree of the numerator is greater than the degree of the denominator. You will not have a vertical or a horizontal asymptote, but we will have an oblique asymptote. You can have a vertical asymptote with a horizontal asymptote, and a vertical asymptote with an oblique asymptote. But, you cannot have a horizontal asymptote and an oblique asymptote at the same time. One or another, but not both. Also, the graph of a rational function will never cross a vertical asymptote, but the graph can cross a horizontal or oblique asymptote.

EXAMPLE 9:

Include all asymptotes, maximums and minimums, if any, where the function is increasing and decreasing, concavity, and any points of inflection, if any.

SOLUTION: x ¹ 3, so there is a vertical asymptote at x = 3. The degree of the numerator is smaller than the degree of the denominator, therefore, there is a horizontal asymptote and it is y = 0.

Now let us determine where the function is increasing and decreasing.

Therefore, the only critical point is x = 3, the vertical asymptote.

(-¥ , 3)

Increasing

(3, ¥ )

Increasing

This function has no local extrema.

Now to determine concavity.

Therefore, the only critical point is x = 3, the vertical asymptote.

(-¥ , 3)

ccu

(3, ¥ )

ccd

Since x = 3 is a vertical asymptote, then it cannot be a point of inflection. Here is the graph of the function. Does it fit the data we have found? Yes it does!

EXAMPLE 10:

Include all asymptotes, maximums and minimums, if any, where the function is increasing and decreasing, concavity, and any points of inflection, if any.

SOLUTION: Let us first simplify this function.

So x = -1 is a removable point of discontinuity and x = 1 is a vertical asymptote. Since the degree of the numerator is the same as the denominator, then there is a horizontal asymptote at y = 2.

Now to determine where the function is increasing and decreasing.

Therefore, the only critical points are x = -1, the removable point of discontinuity, and x = 1, the vertical asymptote. We must include the removable point of discontinuity in the list of critical points because it is a point where the function is undefined.

(-¥ , -1)

Decreasing

(-1, 1)

Decreasing

(1, ¥ )

Decreasing

There are no extreme values for this function. Now to investigate the continuity of this function.

Again, the critical points are x = -1 and x = 1.

(-¥ , -1)

ccd

(-1, 1)

ccd

(1, ¥ )

ccu

Again, x = 1 is not a point of inflection since it is a vertical asymptote. Does the information that we have found fit the graph? Yes it does!

EXAMPLE 11:

Include all asymptotes, maximums and minimums, if any, where the function is increasing and decreasing, concavity, and any points of inflection, if any.

SOLUTION: The vertical asymptote for this function is x = 1.

Since the degree of the numerator is greater than the degree of the denominator, this function will have an oblique asymptote. To find this asymptote, we will perform long division. The oblique asymptote will be the

quotient, and always write it in the form of an equation.

The oblique asymptote is y = x. (This equation will be graphed in red in the graph.)

Now to determine where the function is increasing and decreasing.

The critical points are x = 0, x = 1, and x = 2.

(-¥ , 0)

Increasing

(0, 1)

Decreasing

(1, 2)

Decreasing

(2, ¥ )

Increasing

This function has a local maximum at x = 0 and f (0) = -1.

This function has a local minimum at x = 2 and f (2) = 3.

Now to determine the concavity of this function.

So the vertical asymptote, x = 1 is the only critical point.

(-¥ , 1)

ccd

(1, ¥ )

ccu

Since x = 1 is a vertical asymptote, then it cannot be a point of inflection. Here is the graph of this function. Determine if it fits the data that we have collected.