Determining Limits Using Algebraic Properties of Limits - AP Calculus AB Study Guide

The Core Idea: Determining Limits Using Algebraic Properties of Limits

While limits can be estimated by analyzing graphs or tables of values, the core of calculus requires precise, exact answers. This topic introduces the fundamental algebraic rules for calculating the exact value of a limit without relying on graphical or numerical estimation. The central idea is that the limit of a complex function can often be found by breaking it down into simpler parts and combining their individual limits using a set of established properties.

These properties allow us to handle sums, differences, products, quotients, and compositions of functions. Crucially, this topic also provides a systematic way to handle situations where simply substituting the value $x=c$ into the function leads to problematic expressions. We will learn to distinguish between a limit that is truly undefined (like a non-zero number divided by zero) and a limit that is "indeterminate" (like $0/0$), which requires further algebraic manipulation, such as factoring or rationalizing, to reveal the true value of the limit.

Key Rules

The following properties allow for the calculation of limits of functions that are combined in various ways. These rules apply provided that the individual limits, ${\lim }_{x\to c}f(x)$ and ${\lim }_{x\to c}g(x)$, exist.

• Limit of a Constant: The limit of a constant function is simply the constant itself.

  $$\underset{x\to c}{\lim }k=k$$

• Limit of a Sum or Difference: The limit of a sum or difference of two functions is the sum or difference of their individual limits.

  $$\underset{x\to c}{\lim }[f(x)\pm g(x)]=\underset{x\to c}{\lim }f(x)\pm \underset{x\to c}{\lim }g(x)$$

• Limit of a Constant Multiple: The limit of a constant multiplied by a function is the constant multiplied by the limit of the function.

  $$\underset{x\to c}{\lim }[k\cdot f(x)]=k\cdot \underset{x\to c}{\lim }f(x)$$

• Limit of a Product: The limit of a product of two functions is the product of their individual limits.

  $$\underset{x\to c}{\lim }[f(x)\cdot g(x)]=\left(\underset{x\to c}{\lim }f(x)\right)\cdot \left(\underset{x\to c}{\lim }g(x)\right)$$

• Limit of a Quotient: The limit of a quotient of two functions is the quotient of their individual limits, provided the limit of the denominator is not zero.

  $$\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\frac{{\lim }_{x\to c}f(x)}{{\lim }_{x\to c}g(x)},\text{provided }\underset{x\to c}{\lim }g(x)\ne 0$$

• Limit of a Composite Function: The limit of a composite function can be found by taking the limit of the inner function and then applying the outer function to that result.

  $$\underset{x\to c}{\lim }g(f(x))=g\left(\underset{x\to c}{\lim }f(x)\right)$$

Understanding Indeterminate vs. Undefined Forms

When attempting to evaluate a limit using direct substitution (i.e., plugging the value $c$ into the function), you may encounter two special forms involving zero. It is critical to distinguish between them.

Undefined Limits: The $\frac{k}{0}$ Form (where $k\ne 0$)

If direct substitution into a quotient $\frac{f(x)}{g(x)}$ results in a non-zero number in the numerator and zero in the denominator, the limit is considered undefined. This situation often corresponds to a vertical asymptote on the graph of the function at $x=c$. The function's values grow without bound (towards $\infty$ or $-\infty$) as $x$ approaches $c$, so a finite limit does not exist. No further algebraic simplification can produce a finite value.

Indeterminate Forms: The $\frac{0}{0}$ Form

If direct substitution results in the form $\frac{0}{0}$, this is called an indeterminate form. This result does not mean the limit is 0, 1, or undefined. Instead, it signals that more investigation is required because the limit may or may not exist. The value of the limit is "hidden" by the fact that both the numerator and denominator are approaching zero simultaneously. To find the limit, you must use algebraic techniques to rewrite the expression in a form where direct substitution is possible. The two primary techniques are:

1. Factoring: Factor the numerator and denominator to find and cancel a common factor that is causing the $\frac{0}{0}$ issue.

2. Rationalizing: If the expression involves a square root, multiply the numerator and denominator by the conjugate of the part containing the root. This often resolves the $\frac{0}{0}$ form.

Core Concepts & Rules

• The first step in evaluating a limit algebraically is always to try direct substitution. If this yields a finite number, that number is the limit.

• The limit of a combination of functions (sum, difference, product, quotient) can be found by applying the limit properties to each part individually.

• The quotient rule for limits is only valid if the limit of the denominator is not zero.

• If direct substitution yields a non-zero number divided by zero ($\frac{k}{0}$), the limit is undefined and does not exist.

• If direct substitution yields the indeterminate form $\frac{0}{0}$, this is a signal to perform algebraic manipulation (such as factoring or rationalizing) to simplify the expression before attempting substitution again.

• For a composite function $g(f(x))$, the limit can be passed inside the outer function: $g({\lim }_{x\to c}f(x))$.

Step-by-Step Example 1: Using Limit Properties

Problem: Evaluate ${\lim }_{x\to 2}(4x^2-3x+5)$.

Step 1: Apply the Sum and Difference Rule.

Break the limit of the polynomial into the limits of its individual terms.

$$\underset{x\to 2}{\lim }(4x^2-3x+5)=\underset{x\to 2}{\lim }(4x^2)-\underset{x\to 2}{\lim }(3x)+\underset{x\to 2}{\lim }(5)$$

Step 2: Apply the Constant Multiple Rule.

Factor out the constant coefficients from the first two terms.

$$=4\cdot \underset{x\to 2}{\lim }(x^2)-3\cdot \underset{x\to 2}{\lim }(x)+\underset{x\to 2}{\lim }(5)$$

Step 3: Evaluate the remaining limits using direct substitution.

The limit of $x^2$ as $x\to 2$ is $2^2=4$. The limit of $x$ as $x\to 2$ is $2$. The limit of a constant, $5$, is $5$.

$$=4\cdot (2^2)-3\cdot (2)+5$$

Step 4: Simplify the expression.

$$=4\cdot (4)-6+5$$

$$=16-6+5$$

$$=15$$

Final Answer:${\lim }_{x\to 2}(4x^2-3x+5)=15$. Note that this is the same result as directly substituting $x=2$ into the original polynomial, which is a valid shortcut for all polynomial functions.

Step-by-Step Example 2: Exam-Style Application (Indeterminate Form)

Problem: Find the value of ${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}$.

Step 1: Attempt Direct Substitution.

Substitute $x=0$ into the expression.

$$\frac{\sqrt{0+9}-3}{0}=\frac{\sqrt{9}-3}{0}=\frac{3-3}{0}=\frac{0}{0}$$

This is an indeterminate form, which means we must perform algebraic manipulation.

Step 2: Choose an Algebraic Technique.

The presence of a square root in the numerator suggests that rationalizing is the appropriate method. We will multiply the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x+9}+3$.

$$\underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$$

Step 3: Simplify the Expression.

Multiply the numerators. Remember that $(a-b)(a+b)=a^2-b^2$.

Numerator: $(\sqrt{x+9}{)}^2-(3{)}^2=(x+9)-9=x$.

Denominator: $x(\sqrt{x+9}+3)$.

The expression becomes:

$$\underset{x\to 0}{\lim }\frac{x}{x(\sqrt{x+9}+3)}$$

Step 4: Cancel the Common Factor.

The factor $x$ in the numerator and denominator is causing the $\frac{0}{0}$ form. We can cancel it, as long as $x\ne 0$. Since a limit only cares about values near$x=0$, not at$x=0$, this cancellation is valid.

$$\underset{x\to 0}{\lim }\frac{1}{\sqrt{x+9}+3}$$

Step 5: Re-evaluate the Limit with Direct Substitution.

Now, substitute $x=0$ into the simplified expression.

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

Final Answer:${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}=\frac{1}{6}$.

Using Your Calculator

The algebraic determination of limits is a purely analytical skill. A calculator cannot perform the symbolic manipulation (like factoring or rationalizing) required to solve these problems. Therefore, you will not use a calculator to find the answer on the exam.

However, you can use your calculator to verify your answer.

To verify the answer from Example 2 (${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}=\frac{1}{6}$):

1. Press the Y= button and enter the function: Y1 = (√(X+9) - 3) / X.

2. Press the GRAPH button to see the function's behavior near $x=0$. You should see a "hole" at $x=0$, but the function values should be approaching a specific y-value from both sides.

3. Use the TABLE feature. Press 2nd then TBLSET. Set TblStart = 0 and ΔTbl to a small number like $0.001$.

4. Press 2nd then TABLE. Observe the y-values for x-values very close to 0 (e.g., at $-0.001$ and $0.001$). You will see that the y-values are very close to $\mathrm{0.1666...}$, which is the decimal representation of $\frac{1}{6}$. This provides strong numerical support for your analytical answer.

AP Exam Quick Hit

Common Question Types

• Direct Calculation of an Indeterminate Form: You will be asked to evaluate a limit that results in $\frac{0}{0}$ upon direct substitution. This is the most common question type for this topic.

  • Example: Find ${\lim }_{x\to 4}\frac{x^2-x-12}{x-4}$. (This requires factoring the numerator).

• Using Properties with Given Information: You will be given the values of limits for two functions, $f$ and $g$, and asked to find the limit of an algebraic combination of them.

  • Example: If ${\lim }_{x\to 1}f(x)=8$ and ${\lim }_{x\to 1}g(x)=-2$, find ${\lim }_{x\to 1}\frac{f(x)}{3g(x)}$.

Common Mistakes

• Stopping at $\frac{0}{0}$: A very common error is to see the $\frac{0}{0}$ form and incorrectly conclude that the limit is 0, 1, or undefined. Remember, $\frac{0}{0}$ is an indeterminate form that signals the need for more algebraic work.

• Confusing $\frac{0}{0}$ with $\frac{k}{0}$: Students may mistakenly try to factor or rationalize an expression like ${\lim }_{x\to 1}\frac{5}{x-1}$. This limit results in the form $\frac{5}{0}$, which is immediately undefined; no further work is needed or possible.

• Algebraic Errors: Simple mistakes in factoring (e.g., $x^2-9=(x-3)(x-3)$) or distributing a negative sign after rationalizing can lead to an incorrect final answer even if the calculus concept is understood.

• Forgetting to Multiply by the Conjugate in Both Numerator and Denominator: When rationalizing, a student might only multiply the numerator by the conjugate, changing the value of the entire expression and leading to an incorrect limit.

• Incorrectly Applying the Quotient Rule: Applying the formula for the limit of a quotient when the limit of the denominator is zero. The rule explicitly states it is not valid in this case.