1.4: Algebraic Manipulations and the Limit Laws

Section Preview

Note to the Instructor (click to expand)

This is the moment where you can really distinguish between the value of a function at a point and the value near to that point. The Locally Equivalent Functions material should really drive the point home that, when Direct Substitution fails us (which it often will), we can lean on a "locally-equivalent" function to evaluate. The "Mathematical Mantra" is introduced, and you should repeat this frequently throughout the course. There is a lot of old Algebra in this section, so you might need to remind students of factoring techniques. Using the piecewise definition of the absolute value function is also a critical topic!

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Absolute Value Functions
  • The Absolute Value Function

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Factoring Techniques
  • Greatest Common Factor
  • Factoring Trinomials
  • Factoring Difference of Squares
  • Factoring Sums and Differences of Cubes
• Simplifying Expressions
  • Adding and Subtracting Rational Expressions
  • Simplifying Compound Rational Expressions

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Evaluate the limit of a function by first performing algebraic manipulations (factoring, conjugate multiplication, combining fractions, etc.).
• Evaluate the limit of a function involving absolute values using the definition of an absolute value function.

Most limits encountered in Calculus cannot be immediately evaluated using the Direct Substitution Property. There is often some bit of required manipulation to transform the function within the limit to something more amenable to our needs. It is this action—manipulating the function within the limit—that is the focus of this section. Before we begin, it is important to introduce (or reintroduce, if you have seen it before) the Mathematical Mantra.

The Mathematical Mantra

The following "mantra" is very helpful throughout all levels of Mathematics, and significantly more so in Calculus.

Perform mathematics in the order you learned mathematics. That is, perform

Arithmetic before Algebra,

Algebra before Trigonometry,

Trigonometry before Precalculus, and

Precalculus before Differential Calculus.

Always be willing to ask yourself, "Is there something I learned previously that can simplify my work here?"

We will reference the Mathematical Mantra throughout the rest of Calculus.

Caution: The Mathematical Mantra is a Rule of Thumb

The Mathematical Mantra is a "rule of thumb" and, as such, there are times when it is not applicable. In such cases, I will endeavor to point out why it does not apply.

Locally Equivalent Functions

As we have seen, we may easily evaluate the limits of polynomials and limits of some (but not all) rational functions using the Direct Substitution Property (for polynomial and rational functions). However, as we saw in the introductory section on limits, it is certainly possible for \(\displaystyle \lim_{x \to a}f(x)\) to exist when \(f(a)\) is undefined. The following theorem allows us to evaluate many limits of this type.

Theorem: Locally Equivalent Functions Have Equivalent Limits

If for all \(x \neq a\) we have \(f(x)=g(x)\) over some open interval containing \(a\), then\[\lim_{x \to a}f(x)=\lim_{x \to a}g(x),\nonumber\]provided these limits exist.

This unlocks our ability to evaluate limits we otherwise would never be able to determine.

To understand this idea better, consider the limit \(\displaystyle \lim_{x \to 1}\dfrac{x^2-1}{x-1}\). The function\[f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}\nonumber\]and the function \(g(x)=x+1\) are identical for all values of \(x \neq 1\). The graphs of these two functions are shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): The graphs of \(f(x)\) and \(g(x)\) are identical for all \(x \neq 1\). Their limits at \(1\) are equal.

We see that\[\lim_{x \to 1}\dfrac{x^2-1}{x-1}=\lim_{x \to 1}\dfrac{(x-1)(x+1)}{x-1}=\lim_{x \to 1}\,(x+1)=2.\nonumber\]The limit has the form \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}\), where \(\displaystyle\lim_{x \to a}f(x)=0\) and \(\displaystyle\lim_{x \to a}g(x)=0\). (In this case, we say that \(\frac{f(x)}{g(x)}\) has the indeterminate form \(\frac{0}{0}\).) When evaluating limits of this type, it is imperative to follow the Mathematical Mantra.

Locally Equivalent Functions - Found by Factoring

We spent years in Algebra learning how to factor, but never were told why (unless you had an excellent instructor). The truth is, factoring is necessary in Calculus. It allows us to create the "possible" from something that seems impossible. All of our previous factoring techniques—factoring differences of squares, sums of cubes, differences of cubes, and trinomials, to name the most popular—will be needed at one point or another. In fact, we might even need to dust off the old Rational Roots Theorem at times!

Example \(\PageIndex{1}\): Evaluating a Limit by Factoring and Canceling

Evaluate \(\displaystyle\lim_{x \to 3}\dfrac{x^2-3x}{2x^2-5x-3}\).

Solution

The function \(f(x)=\frac{x^2-3x}{2x^2-5x-3}\) is undefined for \(x=3\). In fact, if we substitute \(3\) into the function we find this limit is indeterminate of the form \(0/0\). This means that we must follow the Mathematical Mantra and perform any Arithmetic, Algebra, and then Trigonometry before attempting to apply Calculus here.

From Algebra, we can see that factoring in both the numerator and denominator is possible.\[\begin{array}{rclcl} \displaystyle \lim_{x \to 3}{\dfrac{x^2-3x}{2x^2-5x-3}} & = & \displaystyle \lim_{x \to 3}{\dfrac{x(x-3)}{(x-3)(2x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 3}{\dfrac{x\cancel{(x-3)}}{\cancel{(x-3)}(2x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 3}{\dfrac{x}{2x+1}} & & \\[8pt] & \overset{D.S.}{=} & \dfrac{3}{7} & \quad & \left(\text{Direct Substitution} \right)\\ \end{array}\nonumber\]

Example \(\PageIndex{1}\) uses the fact that \(f(x) =\frac{x^2-3x}{2x^2-5x-3}\) is identical to the function \(y = \frac{x}{2x + 1}\) as long as \(x \neq 3\). Therefore, around \(x = 3\), but not at \(x = 3\), these two functions are locally equivalent. Remember, a limit as \(x \to 3\) is asking us to investigate the function as \(x\) gets as close as we want to \(3\), without ever becoming \(3\).

Caution: Limits Require... Limits

A widespread mistake is not using limit notation when working with limits. Be aware that limit notation is incredibly important in Calculus and must be used until you finish evaluating the limit. Example \(\PageIndex{1}\) illustrates the proper use of limit notation. The following, however, would be absolutely incorrect:\[\begin{array}{rcl} \displaystyle \lim_{x \to 3}{\dfrac{x^2-3x}{2x^2-5x-3}} & = & \displaystyle \lim_{x \to 3}{\dfrac{x(x-3)}{(x-3)(2x+1)}} \\[8pt] & = & \dfrac{x\cancel{(x-3)}}{\cancel{(x-3)}(2x+1)} \\[8pt] & = & \dfrac{x}{2x+1} \\[8pt] & \overset{D.S.}{=} & \frac{3}{7} \\ \end{array}\nonumber\]

Checkpoint \(\PageIndex{1}\)

Evaluate \(\displaystyle \lim_{x \to -3}\frac{x^2+4x+3}{x^2-9}\).

Answer

\(\dfrac{1}{3}\)

Locally Equivalent Functions - Found by Conjugate Multiplication

Another manipulation that is commonly needed in Calculus can occur when your limit involves a sum or difference of expressions where at least one is a radical. We need to reach into our bag of Algebra tools to remember what to do in these situations.

Example \(\PageIndex{2}\): Evaluating a Limit by Multiplying by a Conjugate

Recall from the previous section that \(\displaystyle \lim_{\theta \to 0} \cos \theta = 1\). Use this to evaluate the following limit.\[\lim_{t \to 0}{\frac{\sqrt{\cos (t)+ 3} - 2}{\cos (t)- 1}}\nonumber\]

Solution

As \(t\) approaches \(0\), both the numerator and denominator of \(f(x)=\frac{\sqrt{\cos (t)+ 3} - 2}{\cos(t)- 1}\) approach \(0\). Therefore, this limit is indeterminate of the form \(0/0\). This means that we must follow the Mathematical Mantra and perform any Arithmetic, Algebra, and then Trigonometry (generally, in that order) before attempting to apply Calculus here.

The numerator, having two terms where one involves a square root, is enticing. Let's try multiplying the numerator and denominator by the conjugate of the numerator.\[\begin{array}{rclcl} \displaystyle \lim_{t \to 0}{\dfrac{\sqrt{\cos (t)+ 3} - 2}{\cos (t)- 1}} & = & \displaystyle \lim_{t \to 0}{\left(\dfrac{\sqrt{\cos (t)+ 3} - 2}{\cos (t)- 1} \cdot \dfrac{\sqrt{\cos (t)+ 3} + 2}{\sqrt{\cos (t)+ 3} + 2}\right)} & & \\[8pt] & = & \displaystyle \lim_{t \to 0}{\left(\dfrac{\cos (t)+ 3 - 4}{\left( \cos (t)- 1 \right) \left( \sqrt{\cos (t)+ 3} + 2\right)}\right)} & \quad &\left(\text{distributing}^* \text{ (see Note below)}\right) \\[8pt] & = & \displaystyle \lim_{t \to 0}{\left(\dfrac{\cos (t)- 1}{\left( \cos (t)- 1 \right) \left( \sqrt{\cos (t)+ 3} + 2\right)}\right)} & & \\[8pt] & = & \displaystyle \lim_{t \to 0}{\left(\dfrac{\cancel{(\cos (t)- 1)}}{\cancel{\left( \cos (t)- 1 \right)} \left( \sqrt{\cos (t)+ 3} + 2\right)}\right)} & & \\[8pt] & = & \displaystyle \lim_{t \to 0}{\left(\dfrac{1}{\sqrt{\cos (t)+ 3} + 2}\right)} & & \\[8pt] & = & \dfrac{\displaystyle \lim_{t \to 0}{1}}{\displaystyle \lim_{t \to 0}{\left(\sqrt{\cos (t)+ 3} + 2\right)}} & \quad & \left( \text{Quotient Law} \right) \\[8pt] & = & \dfrac{1}{\displaystyle \lim_{t \to 0}{\sqrt{\cos (t)+ 3}} + \lim_{t \to 0}{2}} & \quad & \left( \text{Sum Law and Basic Limit Laws} \right) \\[8pt] & = & \dfrac{1}{\sqrt{\displaystyle \lim_{t \to 0}{\left(\cos (t)+ 3\right)}} + 2} & \quad & \left( \text{Root Law and Basic Limit Laws} \right) \\[8pt] & = & \dfrac{1}{\sqrt{\displaystyle \lim_{t \to 0}{\cos (t)} + \displaystyle \lim_{t \to 0}{3}} + 2} & \quad & \left( \text{Sum Law} \right) \\[8pt] & = & \dfrac{1}{\sqrt{1 + 3} + 2} & \quad & \left( \text{Basic Limit Laws} \right) \\[8pt] & = & \dfrac{1}{4} & & \\ \end{array}\nonumber\]

Caution: Conjugate Multiplication and Distribution

When multiplying the numerator and denominator of an expression by a conjugate (of either the numerator or the denominator), it is advisable to only perform distribution on the conjugate factors.

For example, consider the following two limits\[\lim_{x \to -10}{\frac{\sqrt{110 + x} - 10}{x + 10}} \qquad \text{and} \qquad \lim_{\theta \to 0}{\frac{\sin^3{(\theta)}}{1 - \cos{(\theta)}}}\nonumber\]Both are indeterminate of the form \(0/0\). Multiplying the numerator and denominator of the first limit by \(\sqrt{110 + x} + 10\) and the numerator and denominator of the second by \(1 + \cos{(\theta)}\), we get\[\lim_{x \to -10}{\left( \frac{\sqrt{110 + x} - 10}{x + 10} \cdot \frac{\sqrt{110 + x} + 10}{\sqrt{110 + x} + 10} \right)} \qquad \text{and} \qquad \lim_{\theta \to 0}{\left(\frac{\sin^3{(\theta)}}{1 - \cos{(\theta)}} \cdot \frac{1 + \cos{(\theta)}}{1 + \cos{(\theta)}}\right)}\nonumber\]By distributing the conjugate factors and leaving the non-conjugate factors as an undistributed product, we get\[\lim_{x \to -10}{\left( \frac{110 + x - 100}{\left(x + 10\right) \left(\sqrt{110 + x} + 10\right)} \right)} \qquad \text{and} \qquad \lim_{\theta \to 0}{\left(\frac{\sin^3{(\theta)} \left( 1 + \cos{(\theta)} \right)}{1 - \cos^2{(\theta)}} \right)}\nonumber\]Simplifying...\[\lim_{x \to -10}{\left( \frac{10 + x}{\left(x + 10\right) \left(\sqrt{110 + x} + 10\right)} \right)} \qquad \text{and} \qquad \lim_{\theta \to 0}{\left(\frac{\sin^3{(\theta)} \left( 1 + \cos{(\theta)} \right)}{\sin^2{(\theta)}} \right)}\nonumber\]... something beautiful occurs!\[\lim_{x \to -10}{\left( \frac{\cancel{(10 + x)}}{\cancel{\left(x + 10\right)} \left(\sqrt{110 + x} + 10\right)} \right)} \overset{D.S.}{=} \frac{1}{20} \qquad \text{and} \qquad \lim_{\theta \to 0}{\left( \sin{(\theta)} \left( 1 + \cos{(\theta)} \right) \right)} \overset{D.S.}{=} 0\nonumber\](Note: I did a little bit of "hand-waving" on the Direct Substitution for the trigonometric expression. Technically, we have not learned the Direct Substitution Property for trigonometric functions. Still, the point of the example here was to showcase when to distribute conjugates and when not to.)

Checkpoint \(\PageIndex{2}\)

Evaluate \(\displaystyle \lim_{x \to 5}\frac{\sqrt{x-1}-2}{x-5}\).

Answer

\(\dfrac{1}{4}\)

Locally Equivalent Functions - Found by Simplifying Compound Rational Expressions

Another common algebraic manipulation that occurs throughout Calculus is the need to simplify a compound rational expression.

Example \(\PageIndex{3}\): Evaluating a Limit by Simplifying a Complex Fraction

Evaluate \(\displaystyle \lim_{x \to 1}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\).

Solution

As \(x \to 1\), the function \(f(x)=\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\) approaches \(0/0\). This means that we must follow the Mathematical Mantra and perform any Arithmetic, Algebra, and then Trigonometry (generally, in this order) before attempting to apply Calculus.

From Algebra, we know that any compound fraction like this should be simplified.\[\begin{array}{rclcl} \displaystyle \lim_{x \to 1}{\dfrac{\frac{1}{x+1}-\frac{1}{2}}{x-1}} & = & \displaystyle \lim_{x \to 1}{\dfrac{2-(x + 1)}{2(x-1)(x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 1}{\dfrac{1 - x}{2(x-1)(x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 1}{\dfrac{-(x - 1)}{2(x-1)(x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 1}{\dfrac{-\cancel{(x - 1)}}{2\cancel{(x-1)}(x+1)}} & & \\[8pt] & = & \displaystyle \lim_{x \to 1}{\dfrac{-1}{2(x+1)}} & \quad & \\[8pt] & \overset{D.S.}{=} & -\dfrac{1}{4} & \quad & \left(\text{Direct Substitution}\right) \\ \end{array}\nonumber\]

Checkpoint \(\PageIndex{3}\)

Evaluate \(\displaystyle \lim_{x \to -3}\frac{\frac{1}{x+2}+1}{x+3}\).

Answer

\(-1\)

Locally Equivalent Functions - Found by Using the Definition of the Absolute Value Function

Knowing the true definition of the absolute value (the piecewise definition) is going to be critical in Calculus. Recall,\[| \blacksquare | = \begin{cases} \blacksquare, & \text{ if }\blacksquare\geq 0\\[6pt] -\blacksquare, & \text{ if }\blacksquare< 0 \\[6pt] \end{cases}\nonumber\]

Example \(\PageIndex{4}\): Evaluating a Limit Involving Absolute Values

Evaluate\[\displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{|3x + 5|}{6x^2 + 7x - 5} }.\nonumber\]

Solution

As \(x \to (-5/3)^-\), both the numerator and denominator of the limit tend to \(0\). Therefore, this limit is indeterminate of the form \(\frac{0}{0}\). Following the Mathematical Mantra, we clean things up first by performing some Algebra.

Our prerequisite factoring skills allow us to factor the denominator to the following:\[6x^2 + 7x - 5 = (3x + 5)(2x - 1).\nonumber\]Therefore,\[\displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{|3x + 5|}{6x^2 + 7x - 5} } = \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{|3x + 5|}{(3x + 5)(2x - 1)} }.\nonumber\]We know that \(|3x + 5|\) is not the same as \(3x + 5\), so we must use the definition of the absolute value to rewrite \(|3x + 5|\).\[|3x + 5| = \begin{cases} 3x + 5 & \text{, if }3x + 5 \geq 0 & \implies x \geq -\dfrac{5}{3} \\ -(3x + 5) & \text{, if }3x + 5 \lt 0 & \implies x \lt -\dfrac{5}{3} \\ \end{cases}\nonumber\]Since \(x\) is approaching \(-5/3\) from the left, \(x \lt -5/3\). Therefore, \(|3x + 5| = -(3x + 5)\). Hence,\[\begin{array}{rclcl} \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{|3x + 5|}{6x^2 + 7x - 5} } & = & \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{|3x + 5|}{(3x + 5)(2x - 1)} } & & \\[8pt] & = & \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{-(3x + 5)}{(3x + 5)(2x - 1)} } & \quad & \left(\text{using the definition of the absolute value}\right) \\[8pt] & = & \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{-\cancel{(3x + 5)}}{\cancel{(3x + 5)}(2x - 1)} } & & \\[8pt] & = & \displaystyle \lim_{x \to (-5/3)^-}{ \dfrac{-1}{2x - 1} } & & \\[8pt] & \overset{\text{D.S.}}{=} & \dfrac{-1}{2(-5/3) - 1} & \quad & \left( \text{Direct Substitution} \right) \\[8pt] & = & \dfrac{3}{13} & & \\ \end{array}\nonumber\]

Interactive Element: The Absolute-Value Jump at \(x=-\frac{5}{3}\)

The following Interactive Element might help you visualize why the two-sided limit fails and why the side of approach decides the answer.

Interact: Drag the red point along the curve toward \(x=-\frac{5}{3}\) — first from the left, then from the right.

Observation: From the left the point climbs toward \(\frac{3}{13}\); from the right it drops toward \(-\frac{3}{13}\). They differ because \(|3x+5|\) switches to \(-(3x+5)\) exactly at \(x=-\frac{5}{3}\), so the left-hand limit is \(\frac{3}{13}\).

Despite having written down the precise definition before Example \(\PageIndex{4}\), a cautionary statement is worthwhile.

Critical Mistake: Not Using the True Definition of the Absolute Value

Example \(\PageIndex{4}\) should remind you of the fact that \(|f(x)|\) is not necessarily equivalent to \(f(x)\). This is a critical prerequisite skill for Calculus and it bears repeating here:\[|f(x)| = \begin{cases} f(x), & \text{if }f(x) \geq 0 \\[6pt] -f(x), & \text{if }f(x) \lt 0 \\[6pt] \end{cases}\nonumber\]Knowing the piecewise definition of the absolute value function and using it to manipulate expressions involving absolute values is a key fundamental component to your success in Calculus. Theoretically, lacking this single prerequisite skill will not prevent you from succeeding in Calculus; however, in practice, a student who cannot master the understanding of this concept rarely (if ever) passes a Calculus course.

Locally Equivalent Functions - Found by Simplifying Rational Expressions

Example \(\PageIndex{5}\) does not fall neatly into any of the patterns established in the previous examples. However, we can still use these same techniques with a little creativity.

Example \(\PageIndex{5}\): Evaluating a Limit When the Limit Laws Do Not Apply

Evaluate\[\displaystyle \lim_{x \to 0}\left(\dfrac{1}{x}+\dfrac{5}{x(x-5)}\right).\nonumber\]

Solution

Both \(1/x\) and \(5/x(x-5)\) fail to have a limit at zero. Since neither function has a zero limit, we cannot apply the Sum Law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that\[\dfrac{1}{x}+\dfrac{5}{x(x-5)}=\dfrac{x-5+5}{x(x-5)}=\dfrac{x}{x(x-5)}.\nonumber\]Thus,\[\lim_{x \to 0}\left(\dfrac{1}{x}+\dfrac{5}{x(x-5)}\right)=\lim_{x \to 0}\dfrac{x}{x(x-5)}=\lim_{x \to 0}\dfrac{1}{x-5}=-\dfrac{1}{5}.\nonumber\]

Checkpoint \(\PageIndex{5}\)

Evaluate \(\displaystyle \lim_{x \to 3}\left(\frac{1}{x-3}-\frac{4}{x^2-2x-3}\right)\).

Answer

\(\dfrac{1}{4}\)

Armed with an ability to use your prerequisite Algebra skills, you should be well on your way to success in Calculus!

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MyOpenMath Advice: Check For Feedback

When grading your uploaded (handwritten) work, your professor might have left feedback. Sometimes this feedback can be as simple as, "Please read the written solution," but it can be as robust as a breakdown of your mistake and how you can avoid such a mistake in the future. To check a past due assignment for any feedback, go to the Grades area in Canvas, click on the assignment in question, and you should be able to see any feedback there. If you cannot see feedback, please let your professor know.

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Success in College: Meeting With Your Professor

Meeting with a faculty member might be intimidating. Students are encouraged to use the following guidelines to make the faculty meeting productive. These strategies were developed by the Retention Office at U. of Toledo (OH) and first appeared in National On-Campus Report's sister publication, Recruitment and Retention in Higher Education.

Step 1: Make an appointment

• Review the syllabus for the professor's office hours and make an appointment.
• If you're uncertain of your professor's office hours, ask.
• If your schedule conflicts with the office hours, tell your professor why you can't make the scheduled office hours. Then tell them about your specific concern.
• Set up a specific time to meet with your professor. Be sure to indicate the amount of time you will need.

Step 2: Establish rapport

• Be on time for your appointment.
• When you arrive, be pleasant, smile, introduce yourself again (include your name and class), and shake hands if appropriate.

Step 3: Present your concern

• Focus on the specific questions you've identified as problematic.
• Have your problem(s) written out so that the professor can see where your difficulties are.

Step 4: Provide background information

• Briefly tell your professor about your high school background and preparation for the course – if relevant to solving the problem.
• Explain the study strategies you've used to understand the material.

Step 5: Redirect for clarification

• If the professor's explanation isn't clear, redirect their attention to the specific point where you became confused.
• "Talk through" your problem.

Step 6: Summarize Resolution of the Problem

• "I was missing this step in... I need to apply this formula..."

Step 7: Thank your Professor

• Ask to come back if necessary.

Armed with this information, think about the following questions:

1. How would you make an appointment to talk to your professor if your schedule prevents you from visiting during office hours? If it is not possible to make such an appointment or your professor does not have office hours, think about another way you can get extra help.
2. Some professors prefer you talk to them using their title and last name. For example, you might say Dr. Rodriguez or Ms. Smith. Others prefer that you use just their first name. When asking your professor a question, what name or title do they prefer that you use?

If you are unsure about the answers to either of these questions, feel free to ask your professor - that's what they are there for!