Learning activity 1.3: Determining limits

Introduction

What exactly is a limit? In everyday life, you may have noticed common phrases like this:

“Don’t drive over the speed limit or you might get a ticket.”

“I’ve reached my limit. I can’t eat one more bite!”

In both of these statements, the word “limit” is used to describe a boundary or furthest extent possible. The mathematical use of the word “limit” is the same. In this activity, you will explore the concept of mathematical limits and discover how finding limits can help to determine the instantaneous rate of change of a function.

Reading Limits

The statement $\underset{h \to 0}{l i m} (4 + h) = 4$ reads:

“The limit as the value of $h$ approaches 0 in the expression ( $4 + h$ ) is 4.”

Over the years, mathematicians have created their own code to condense a statement and to use a more universal language. Hence, even though written as $\underset{h \to 0}{l i m} (4 + h) = 4$ it is read as:

“The limit as $h$ approaches 0 of $(4 + h)$ is 4.”

Your Turn

Try it!

Try reading the following statements. English sentences are available after each statement for comparison.

a. $\underset{h \to 5}{l i m} (6 - h) = 1$

Sentence:

The statement $\underset{h \to 5}{l i m} (6 - h) = 1$ is read as “The limit as h approaches 5 of $(6 - h)$ is 1”.

b. $\underset{x \to \infty}{l i m} (4 + x) = \infty$

Sentence:

The statement $\underset{x \to \infty}{l i m} (4 + x)$ is read as “The limit as x approaches infinity of (4+x) is infinity”.

In limits we are going to be dealing with really large and really small numbers. We will use the symbol for infinity ( $\infty$ ) to denote the really large number and we will use zero (0) to denote the really small number.

Here are some basic conclusions to consider:

Basic Rules for Evaluation of Limits

Rule No. 1:

$limit = \frac{constant}{infinity} = 0$

If you are dividing a constant value by a really large number, then the result of that division will be a number so small we consider it to be zero.

Example: $\frac{7}{\infty} = 0$

Rule No. 2:

$limit = \frac{infinity}{constant} = infinity$

If you are dividing a really large number by a constant value, then the result will be a really large number.

Example $\frac{\infty}{7} = \infty$

Rule No. 3:

$limit = \frac{constant}{0} = infinity$

If you are dividing a constant value by a really small number, then the result will be a really large number.

Example: $limit = \frac{7}{0} = \infty$

It seems in this expression that we are considering the ability to divide by zero which we know by all mathematical standards is not allowed. You need to notice that even though we are using zero, we are referring to a really small number.

Rule No. 4:

$\frac{infinity}{infinity}$

If you have a really large number divided by another really large number, you cannot make any conclusions about the value of the expression as you have no real idea what the actual numbers are or which number is actually bigger. All you know for sure is that both numbers are really large.

Therefore no valid conclusion can be made. You will need to simplify or modify the expression before a conclusion can be made. We will review examples of simplifying and modifying later in this learning activity.

Rule No. 5:

$\frac{zero}{zero} = \frac{0}{0}$

Similarly, in this case you know both numbers are really small but have no idea of their real values.

Evaluating Limits

With all these rules in place, we are now ready to begin evaluating limits.

There are various methods and techniques to evaluate limits. All involve using the basic simplifying procedures you mastered in previous mathematics courses.

Let’s review some examples of different approaches and solution methods. Then you can try some on your own.

1. Straight Substitution

This method just substitutes the value for x into the expression to find the value.

Example:

Find $\underset{x \to 5}{l i m} (\frac{x^{2} - 2 x}{4 - x})$ .

$\underset{x \to 5}{l i m} (\frac{x^{2} - 2 x}{4 - x}) = \frac{(5)^{2} - 2 (5)}{4 - 5} = \frac{25 - 10}{- 1} = \frac{15}{- 1} = - 15$

$\underset{x \to 5}{l i m} (\frac{x^{2} - 2 x}{4 - x}) = - 15$

When the substitution method yields an inconclusive statement, you need to try other approaches. Let’s look at some of those.

2. Factoring

This method uses factoring to simplify rational functions.

Example: Find $\underset{x \to - 3}{l i m} (\frac{x^{2} + 7 x + 12}{x + 3})$ .

Try the substitution method.

$\underset{x \to - 3}{l i m} (\frac{x^{2} + 7 x + 12}{x + 3}) = \frac{(- 3)^{2} + 7 (- 3) + 12}{- 3 + 3} = \frac{0}{0}$

This yields an inconclusive statement.

We will need to simplify this expression.

Factor the quadratic and simplify.

$\underset{x \to - 3}{l i m} (\frac{x^{2} + 7 x + 12}{x + 3}) = \underset{x \to - 3}{l i m} \frac{(x + 3) (x + 4)}{x + 3} = \underset{x \to - 3}{l i m} (x + 4)$

Now we can retry substitution.

$\underset{x \to - 3}{l i m} (x + 4) = - 3 + 4 = 1$

$\underset{x \to - 3}{l i m} (\frac{x^{2} + 7 x + 12}{x + 3}) = 1$

3. Dividing Everything

This method divides every term by the variable with the highest power.

Example: Find $\underset{x \to \infty}{l i m} (\frac{5 x^{3} + 3 x^{2} + 1}{6 x^{3} + x^{2} - 3 x})$ .

Try substitution.

$\underset{x \to \infty}{l i m} (\frac{5 x^{3} + 3 x^{2} + 1}{6 x^{3} + x^{2} - 3 x}) = \frac{5 (\infty)^{3} + 3 (\infty)^{2} + 1}{6 (\infty)^{3} + (\infty)^{2} - 3 (\infty)} = \frac{\infty}{\infty}$

This yields an inconclusive statement.

The expressions do not factor.

Since the highest exponent of $x$ is 3, divide all terms by $x^{3}$ .

$\underset{x \to \infty}{l i m} (\frac{5 x^{3} + 3 x^{2} + 1}{6 x^{3} + x^{2} - 3 x})$

$= \underset{x \to \infty}{l i m} (\frac{\frac{5 x^{3}}{x^{3}} + \frac{3 x^{2}}{x^{3}} + \frac{1}{x^{3}}}{\frac{6 x^{3}}{x^{3}} + \frac{x^{2}}{x^{3}} - \frac{3 x}{x^{3}}}) = \underset{x \to \infty}{l i m} (\frac{5 + \frac{3}{x} + \frac{1}{x^{3}}}{6 + \frac{1}{x} - \frac{3}{x^{2}}}) = \frac{5 + \frac{3}{\infty} + \frac{1}{\infty}}{6 + \frac{1}{\infty} - \frac{3}{\infty}} = \frac{5 + 0 + 0}{6 + 0 - 0} = \frac{5}{6}$

$\underset{x \to \infty}{l i m} (\frac{5 x^{3} + 3 x^{2} + 1}{6 x^{3} + x^{2} - 3 x}) = \frac{5}{6}$

4. Rationalizing

This method rationalizes either the numerator or denominator by multiplying both the numerator and denominator by the conjugate of the root function.

Example: Find $\underset{h \to 0}{l i m} \frac{\sqrt{9 + h} - 3}{h}$ .

Try substitution.

$\underset{h \to 0}{l i m} \frac{\sqrt{9 + h} - 3}{h} = \frac{\sqrt{9 + 0} - 3}{0} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$

This yields an inconclusive statement.

Rationalize the numerator by multiply the numerator and denominator by the conjugate $(\sqrt{9 + h} + 3)$ .

$= \underset{h \to 0}{l i m} \frac{\sqrt{9 + h} - 3}{h} = \underset{h \to 0}{l i m} \frac{(\sqrt{9 + h} - 3) (\sqrt{9 + h} + 3)}{h (\sqrt{9 + h} + 3)} = \lim \frac{(9 + h) + 3 \sqrt{9 + h} - 3 \sqrt{9 + h} - 9}{h (\sqrt{9 + h} + 3)} = \underset{h \to 0}{l i m} \frac{(9 + h) - 9}{h (\sqrt{9 + h} + 3)} = \underset{h \to 0}{l i m} \frac{h}{h (\sqrt{9 + h} + 3)}$

Divide the $h$ ’s.

$= \underset{h \to 0}{l i m} \frac{1}{\sqrt{9 + h} + 3}$

Now substitute $h = 0$

$= \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

$\underset{h \to 0}{l i m} \frac{\sqrt{9 + h} - 3}{h} = \frac{1}{6}$

Check Your Understanding

Notebook

Here are some questions for you to try in your notebook.

Answers are available for comparison when you are ready. Please review the solutions to ensure you are following the correct format.

Find the following limits. Both answers and solutions are available after each question.

1. $\underset{x \to 3}{l i m} \frac{x - 3}{x + 3}$

Try substitution.

$\underset{x \to 3}{l i m} \frac{x - 3}{x + 3} = \frac{3 - 3}{3 + 3} = \frac{0}{6} = 0$

$\underset{x \to 3}{l i m} \frac{x - 3}{x + 3} = 0$

2. $\underset{x \to 0}{l i m} \frac{x^{2} + 2 x - 3}{x^{2} + 2}$

Try substitution.

$\underset{x \to 0}{l i m} \frac{x^{2} + 2 x - 3}{x^{2} + 2}$

$= \frac{(0)^{2} + 2 (0) - 3}{(0)^{2} + 2}$

$= \frac{0 + 0 - 3}{0 + 2}$

$= \frac{- 3}{2}$

$\underset{x \to 0}{l i m} \frac{x^{2} + 2 x - 3}{x^{2} + 2} = - \frac{3}{2}$

3. $\underset{x \to - 2}{l i m} \frac{2 x^{2} + 5 x + 2}{x^{2} - 2 x - 8}$

Try substitution.

$\underset{x \to - 2}{l i m} \frac{2 x^{2} + 5 x + 2}{x^{2} - 2 x - 8}$

$= \frac{2 (- 2)^{2} + 5 (- 2) + 2}{(- 2)^{2} - 2 (- 2) - 8} = \frac{0}{0}$

This yields an inconclusive statement.

Try factoring.

$\underset{x \to - 2}{l i m} \frac{2 x^{2} + 5 x + 2}{x^{2} - 2 x - 8}$

$= \underset{x \to - 2}{l i m} \frac{(2 x + 1) (x + 2)}{(x - 4) (x + 2)}$

$= \underset{x \to - 2}{l i m} \frac{2 x + 1}{x - 4}$

$= \frac{2 (- 2) + 1}{- 2 - 4}$

$= \frac{- 3}{- 6}$

$= \frac{1}{2}$

$\underset{x \to - 2}{l i m} \frac{2 x^{2} + 5 x + 2}{x^{2} - 2 x - 8} = \frac{1}{2}$

4. $\underset{x \to \infty}{l i m} 1 3^{- x}$

Try substitution.

$\underset{x \to \infty}{l i m} 1 3^{- x} = 1 3^{- \infty} = \frac{1}{1 3^{\infty}} = \frac{1}{\infty} = 0$

$\underset{x \to \infty}{l i m} 1 3^{- x} = 0$

5. $\underset{x \to 4}{l i m} \frac{x - 4}{\sqrt{x} - 2}$

Try substitution.

$\underset{x \to 4}{l i m} \frac{x - 4}{\sqrt{x} - 2} = \frac{4 - 4}{\sqrt{4} - 2} = \frac{0}{2 - 2} = \frac{0}{0}$

This yields an inconclusive statement.

Try rationalizing the denominator by multiplying by the conjugate $(\sqrt{x} + 2)$ .

$\underset{x \to 4}{l i m} \frac{x - 4}{\sqrt{x} - 2} = \underset{x \to 4}{l i m} \frac{(x - 4) (\sqrt{x} + 2)}{(\sqrt{x} - 2) (\sqrt{x} + 2)}$

$= \underset{x \to 4}{l i m} \frac{(x - 4) (\sqrt{x} + 2)}{x - 4}$

$= \underset{x \to 4}{l i m} (\sqrt{x} + 2)$

$= \sqrt{4} + 2$

$= 2 + 2 = 4$

$\underset{x \to 4}{l i m} \frac{x - 4}{\sqrt{x} - 2} = 4$

6. $\underset{x \to 1}{l i m} \frac{x^{3} - 1}{x^{3} - x^{2} - 4 x + 4}$

Try substitution.

$\underset{x \to 1}{l i m} \frac{x^{3} - 1}{x^{3} - x^{2} - 4 x + 4} = \frac{(1)^{3} - 1}{(1)^{3} - (1)^{2} - 4 (1) + 4} = \frac{0}{0}$

This yields an inconclusive statement.

Try factoring

$\underset{x \to 1}{l i m} \frac{x^{3} - 1}{x^{3} - x^{2} - 4 x + 4} = \underset{x \to 1}{l i m} \frac{(x - 1) (x^{2} + x + 1)}{(x - 1) (x + 2) (x - 2)}$

$= \underset{x \to 1}{l i m} \frac{x^{2} + x + 1}{(x + 2) (x - 2)}$

$= \frac{(1)^{2} + (1) + 1}{(1 + 2) (1 - 2)} = \frac{3}{- 3} = - 1$

$\underset{x \to 1}{l i m} \frac{x^{3} - 1}{x^{3} - x^{2} - 4 x + 4} = - 1$

7. $\underset{x \to 2}{l i m} \frac{x^{4} - 1}{x^{4} + 2 x^{3} - x^{2} - 2 x}$

Try substitution.

$\underset{x \to 2}{l i m} \frac{x^{4} - 1}{x^{4} + 2 x^{3} - x^{2} - 2 x} = \frac{(2)^{4} - 1}{(2)^{4} + 2 (2)^{3} - (2)^{2} - 2 (2)} = \frac{15}{24} = \frac{5}{8}$

$\underset{x \to 2}{l i m} \frac{x^{4} - 1}{x^{4} + 2 x^{3} - x^{2} - 2 x} = \frac{5}{8}$

Technology

Using a graphing application of your choice, graph each of the functions in the Check Your Understanding. Check out the $y$ value of the function when the $x$ value approaches the value given in the limit.

What do you notice?

Join the discussion

Create two new limit problems that require some of the problem-solving techniques other than substitution. Share them in the forum.

Reply to two other problems posted by your peers with your solution and a graph to support your response.

Conclusion

You have now completed Learning Activity 1.3. If you worked through all the Examples and Check Your Understanding questions, you should feel comfortable with:

reading limit expressions
finding the limit of a function

You are now ready to:

submit your Unit 1: Rates of Change Learning Log entries. It will be teacher marked for feedback.

Next Steps

In Learning Activity 1.4, you will learn how limits can be useful in determining the instantaneous rate of change of a function at a specific point along a curve.

Introduction

Reading Limits

Your Turn

Try it!

Basic Rules for Evaluation of Limits

Evaluating Limits

1. Straight Substitution

2. Factoring

Step 1

Step 2

Step 3

3. Dividing Everything

Step 1

Step 2

4. Rationalizing

Step 1

Step 2

Step 3

Step 4

Check Your Understanding

Notebook

step 1

step 2

step 1

step 2

step 1

step 2

Technology

Join the discussion

Conclusion

Next Steps