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

The Core Idea: Determining Limits Using Algebraic Properties of Limits

The fundamental concept of a limit involves understanding the behavior of a function as its input approaches a certain value. While sometimes this can be found by direct substitution, we often encounter situations where this method fails, yielding an indeterminate form such as $0/0$. These forms do not mean the limit is nonexistent or undefined; rather, they signal that the expression in its current form is insufficient to determine the limit's value.

This topic introduces a systematic, analytical approach to resolving such ambiguities. By leveraging a set of algebraic properties, we can manipulate and rewrite complex functions. These properties allow us to break down limits of sums, differences, products, and quotients into the limits of their simpler component parts. The core task is to use algebraic techniques—such as factoring, canceling, or rationalizing—to transform an expression from an indeterminate form into one where the limit becomes clear, allowing us to find a precise numerical value.

Key Rules: Algebraic Properties of Limits

The following properties apply for any functions $f(x)$ and $g(x)$, provided that ${\lim }_{x\to c}f(x)=L$ and ${\lim }_{x\to c}g(x)=M$ both exist.

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

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

2. The Limit of a Sum: The limit of a sum is the sum of the limits.

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

3. The Limit of a Difference: The limit of a difference is the difference of the limits.

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

4. The Limit of a Constant Multiple: The limit of a constant times a function is the constant times the limit of the function.

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

5. The Limit of a Product: The limit of a product is the product of the limits.

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

6. The Limit of a Quotient: The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.

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

7. The Limit of a Composite Function: If a function $f$ is continuous at the value $M={\lim }_{x\to c}g(x)$, then the limit can be passed through the outer function.

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

Understanding Indeterminate Forms

An indeterminate form is an expression that arises from an initial attempt to evaluate a limit by direct substitution, but which does not have a well-defined value. The presence of an indeterminate form is a crucial signal that the limit cannot be determined from the expression as written and that further algebraic manipulation is required.

The most common indeterminate forms encountered are:

• $\frac{0}{0}$

• $\frac{\infty }{\infty }$

• $\infty -\infty$

• $0\cdot \infty$

It is a critical error to conclude that a limit "equals" $0/0$ or that it does not exist simply because an indeterminate form appears. For example, the Quotient Rule for limits explicitly states that the limit of the denominator cannot be zero. If direct substitution leads to a zero in the denominator, the Quotient Rule cannot be applied. If it leads to $0/0$, you must algebraically rewrite the function to find an equivalent expression (for all $x\ne c$) where the limit can be determined.

Core Concepts & Rules

• Limit Properties: The limits of sums, differences, products, and constant multiples can be calculated by performing those same operations on the individual limits, assuming those individual limits exist.

• Quotient Rule Condition: The limit of a quotient of functions is the quotient of their limits, but this rule is only valid if the limit of the function in the denominator is not zero.

• Indeterminate Form Signal: Encountering an indeterminate form like $0/0$ after direct substitution means that more analytical work is necessary. It is not the final answer.

• Algebraic Manipulation: The primary strategy for resolving indeterminate forms is to use algebraic techniques. This includes factoring and canceling common factors, multiplying by a conjugate to rationalize a numerator or denominator, or simplifying complex fractions.

• Composite Function Limits: To find the limit of a composite function $f(g(x))$, you can first find the limit of the inner function, $g(x)$, and then evaluate the outer function, $f(x)$, at that resulting value. This "shortcut" is only valid if the outer function $f$ is continuous at the limit of the inner function.

Step-by-Step Example 1: Resolving 0/0 by Factoring

Problem: Find the value of ${\lim }_{x\to 3}\frac{x^2-x-6}{x-3}$.

Step 1: Attempt Direct Substitution

First, substitute $x=3$ into the expression to see if the limit can be found directly.

$$\frac{(3{)}^2-(3)-6}{3-3}=\frac{9-3-6}{0}=\frac{0}{0}$$

This is an indeterminate form. This tells us that we cannot determine the limit from the function as written and must perform algebraic manipulation.

Step 2: Algebraic Manipulation (Factoring)

The presence of the indeterminate form suggests that the numerator and denominator may share a common factor. We will factor the quadratic in the numerator.

$$x^2-x-6=(x-3)(x+2)$$

Now, rewrite the entire limit expression with the factored numerator.

$$\underset{x\to 3}{\lim }\frac{(x-3)(x+2)}{x-3}$$

Step 3: Simplify the Expression

Since the limit is concerned with values of $x$approaching 3, but not equal to 3, we know that $x-3\ne 0$. Therefore, we can safely cancel the $(x-3)$ term from the numerator and denominator.

$$\underset{x\to 3}{\lim }(x+2)$$

Step 4: Re-evaluate the Limit

Now we have a simplified expression. We can use direct substitution on this new expression.

$$3+2=5$$

Step 5: Final Answer

The value of the limit is 5.

$$\underset{x\to 3}{\lim }\frac{x^2-x-6}{x-3}=5$$

Step-by-Step Example 2: Using Limit Properties with Given Limits

Problem: Let $f$ and $g$ be functions such that ${\lim }_{x\to -1}f(x)=6$ and ${\lim }_{x\to -1}g(x)=-3$. Find ${\lim }_{x\to -1}\frac{g(x)}{f(x)-2g(x)}$.

Step 1: Apply the Quotient Rule for Limits

We begin by applying the property for the limit of a quotient.

$$\underset{x\to -1}{\lim }\frac{g(x)}{f(x)-2g(x)}=\frac{{\lim }_{x\to -1}g(x)}{{\lim }_{x\to -1}(f(x)-2g(x))}$$

This step is only valid if the limit of the denominator is not zero. We must verify this.

Step 2: Evaluate the Limit of the Denominator

Using the Difference and Constant Multiple rules, we evaluate the denominator's limit.

$$\underset{x\to -1}{\lim }(f(x)-2g(x))=\underset{x\to -1}{\lim }f(x)-\underset{x\to -1}{\lim }(2g(x))$$

$$=\underset{x\to -1}{\lim }f(x)-2\cdot \underset{x\to -1}{\lim }g(x)$$

Now, substitute the given limit values.

$$=6-2(-3)=6+6=12$$

Step 3: Verify the Quotient Rule Condition

Since the limit of the denominator is 12, which is not 0, our application of the Quotient Rule in Step 1 was valid.

Step 4: Substitute All Given Values and Finalize

Now we can substitute the known limits into the expression from Step 1.

$$\frac{{\lim }_{x\to -1}g(x)}{{\lim }_{x\to -1}(f(x)-2g(x))}=\frac{-3}{12}$$

Step 5: Simplify the Final Answer

$$\frac{-3}{12}=-\frac{1}{4}$$

The value of the limit is $-1/4$.

Using Your Calculator

The algebraic properties of limits are analytical tools, and problems on the AP Exam will require you to show the algebraic steps (like factoring or applying properties). A calculator cannot perform these symbolic manipulations for you.

However, a graphing calculator is an excellent tool for checking your answer or building intuition about a limit.

To check the result of Example 1, ${\lim }_{x\to 3}\frac{x^2-x-6}{x-3}$:

1. Graph the Function: Enter Y1 = (X^2 - X - 6) / (X - 3) into your calculator's graphing editor. When you graph it, it will look like the line $y=x+2$, but the calculator will not show the hole at $x=3$.

2. Use the Table Feature:

   • Go to TBLSET (Table Setup).

   • Set TblStart to 3 (the value $x$ is approaching).

   • Set ΔTbl (Delta Table) to a very small number, like $0.001$.

   • View the TABLE. You will see an ERROR for $x=3$, but you can see the $y$-values for $x$ just above and below 3. For example, at $x=2.999$, $y=4.999$, and at $x=3.001$, $y=5.001$. This provides strong numerical evidence that the limit is 5.

3. Use the Trace Feature:

   • Graph the function.

   • Press TRACE. Enter values very close to 3, such as $2.99$ or $3.01$, and observe the corresponding $y$-values, which will be very close to 5.

AP Exam Quick Hit

Common Question Types

• Resolving $0/0$ with a Rational Function: You will be asked to find the limit of a rational function that initially evaluates to $0/0$. The required technique is typically factoring and canceling.

  • Example: Find ${\lim }_{x\to -5}\frac{x^2+3x-10}{x+5}$.

• Applying Limit Properties to Given Abstract Functions: You will be given the limits of two functions, $f(x)$ and $g(x)$, at a point and asked to find the limit of an algebraic combination of them.

  • Example: If ${\lim }_{x\to 2}f(x)=4$ and ${\lim }_{x\to 2}g(x)=-1$, find ${\lim }_{x\to 2}[3f(x)\cdot g(x)]$.

• Resolving $0/0$ with Radicals: You will be asked to find the limit of a function involving a square root that evaluates to $0/0$. The required technique is to multiply the numerator and denominator by the conjugate of the expression with the radical.

  • Example: Find ${\lim }_{h\to 0}\frac{\sqrt{4+h}-2}{h}$.

Common Mistakes

• Stopping at $0/0$: A very common mistake is to see the indeterminate form $0/0$ and incorrectly conclude that the limit Does Not Exist (DNE) or is equal to 0. Remember, $0/0$ is a signal to do more work, not a final answer.

• Improper Use of the Quotient Rule: Applying the quotient rule $\lim \frac{f(x)}{g(x)}=\frac{\lim f(x)}{\lim g(x)}$ when $\lim g(x)=0$. This rule is explicitly not valid in that case. You must use other methods.

• Algebraic Errors: Simple mistakes in factoring polynomials, distributing a negative sign, or multiplying by a conjugate are frequent sources of incorrect answers even when the correct calculus process is chosen.

• Forgetting to Write `lim$: In free-response questions, dropping the ${\lim }_{x\to c}$ notation in the middle of your algebraic steps is a linkage error. You must carry the limit notation through each step of your work until you perform the final substitution.

  • Incorrect: ${\lim }_{x\to 3}\frac{(x-3)(x+2)}{x-3}=(x+2)=3+2=5$

  • Correct: ${\lim }_{x\to 3}\frac{(x-3)(x+2)}{x-3}={\lim }_{x\to 3}(x+2)=3+2=5$