You can recognize the limits by what happens when you substitute the value x approaches into the expression. If it gives 0/0, there is algebra that you can do to find the exact value of the limit.

In the first two examples, the expression may be factored and simplified…then you can substitute the value for x.

In the next two examples, the fractions in the numerator must be combined before the fraction may be simplified.

The next two examples are designed to throw you off. When you substitute the value into the expression, you do not get 0/0. This means you need to use a table or graph to get the limit.

The next two examples show how to rationalize the numerator to do a limit.

This example may be done two different ways as the next two boards demonstrate.

One of the problems on the homework gave you three points on a line graph, (1905, 1024), (1955, 240), (2005, 1141). In these ordered pairs, the x value is the year and the y value is the number of immigrants (in thousands) to a large country.

Find the average rate of change in immigration from 1905 to 1955 in immigrants per year.

Find the average rate of change in immigration from 1955 to 2005 in immigrants per year.

Find the average rate of change in immigration from 1905 to 2005 in immigrants per year.

This problem illustrates the two ways that you can work in the “thousands” in the data to give immigrants per year instead of thousands of immigrants per year.

Problem 1 Find the average rate of change of over [2, 4] to four decimal places.

Problem 2 Find the average rate of change of over [1, 3] to four decimal places.

When you calculate the rate to four decimal places, you should write the numbers in the quotient to FIVE decimal places to make sure there are no rounding errors.

You may have notice that there are two ways to use algebra to compute limits at infinity. Let’s look at two possible strategies for computing a limit at infinity for a rational function, .

In the board above, I divided each term in the rational expression by x to the highest power that appears anywhere in the expression. The denominator get very small as the top approaches 2, so the entire expression must grow very big. Since the bottom is also negative, the expression grows big and is negative. Now let’s change that strategy a bit.

In the board above, I divide by x to the largest power that appears in the denominator. The bottom then gets closer and closer to 3. But the top contains 2x which grow more and more negative as x approaches . So the entire expression grows more and more negative.

Either strategy gives the correct answer as long as you interpret what is going on in the numerator and denominator correctly.