Determining Limits Using Algebraic Manipulation - AP Calculus BC Study Guide

The Core Idea: Determining Limits Using Algebraic Manipulation

When evaluating limits, the first approach is always direct substitution. However, in many cases, substituting the value $c$ into the function $f(x)$ results in an "indeterminate form" such as $$\frac{0}{0}$$. This outcome does not mean the limit is undefined or does not exist. Instead, it signals that the function has a point of discontinuity (like a hole in the graph) at $x=c$, and more investigation is required to determine the function's behavior as $x$ approaches $c$.

This topic focuses on the essential algebraic techniques used to resolve these indeterminate forms. By manipulating the function's expression—through methods like factoring, multiplying by a conjugate, or simplifying complex fractions—we can rewrite it into an equivalent form that is defined at $x=c$. This new expression is identical to the original for all values except at $x=c$ itself. This process allows us to eliminate the source of the indeterminacy and then use direct substitution on the simplified function to find the true value of the limit.

Key Algebraic Techniques

When direct substitution results in an indeterminate form, one of the following algebraic techniques can be used to rewrite the expression and evaluate the limit.

• Factoring and Canceling: This technique is used when the numerator and/or denominator are polynomials that can be factored. The goal is to find a common factor that is causing the $$\frac{0}{0}$$ form and cancel it.

  • Example Form: $$\underset{x\to c}{\lim }\frac{P(x)}{Q(x)}$$ where $P(c)=0$ and $Q(c)=0$.

• Multiplying by the Conjugate: This technique is primarily used when the expression involves a square root, typically in the form $$\sqrt{a}-b$$ or $$\sqrt{a}+b$$. Multiplying both the numerator and the denominator by the conjugate (e.g., $$\sqrt{a}+b$$ is the conjugate of $$\sqrt{a}-b$$) can help eliminate the radical from the part of the fraction causing the indeterminacy.

  • Example Form: $$\underset{x\to c}{\lim }\frac{\sqrt{f(x)}-L}{g(x)}$$ where $$\sqrt{f(c)}-L=0$$ and g(c)=0`. * **Rewriting a Complex Fraction:** This technique is used when the function is a fraction that contains other fractions in its numerator, denominator, or both. The strategy is to find a common denominator for the smaller fractions and simplify the entire expression into a single, standard fraction. This often reveals a common factor that can be canceled. * Example Form: Formula[9] where $f(c)=L and $g(c)=0$.

Understanding Indeterminate Forms

The central challenge addressed by algebraic manipulation is the indeterminate form $$\frac{0}{0}$$. It is crucial to understand what this form signifies.

An indeterminate form like $$\frac{0}{0}$$ or $$\frac{\infty }{\infty }$$ means the limit's value cannot be determined from the form of the expression alone. It indicates a competition between the numerator and the denominator as they both approach zero (or infinity). The limit could be any real number, it could be $$\infty$$ or $$-\infty$$, or it might not exist at all. The purpose of algebraic manipulation is to resolve this "competition" and find the definitive result.

For example, if $$\underset{x\to c}{\lim }f(x)=\frac{0}{0}$$, it suggests there is a common factor of $(x-c)$ in both the numerator and the denominator. Geometrically, this corresponds to a "hole" in the graph of $f(x)$ at $x=c$. The limit is the y-coordinate of this hole. Our algebraic techniques are designed to "patch" this hole by canceling the $(x-c)$ factor, allowing us to evaluate the function's value at that point.

It is critical to distinguish an indeterminate form from a form that indicates a vertical asymptote. If direct substitution yields $$\frac{k}{0}$$ where $k$ is a non-zero constant, the limit does not exist and typically approaches $$\infty$$ or $$-\infty$$. This is not an indeterminate form, and the algebraic techniques discussed here do not apply.

Core Concepts & Rules

• Always begin evaluating a limit by attempting direct substitution.

• If direct substitution results in a determinate value (e.g., a real number), that value is the limit.

• If direct substitution results in an indeterminate form, such as $$\frac{0}{0}$$ or $$\frac{\infty }{\infty }$$, the limit's value is not yet known. Further analysis is required.

• To resolve a $$\frac{0}{0}$$ indeterminate form, use an appropriate algebraic technique to rewrite the function's expression.

• Factoring: Use for rational functions. Factor the numerator and denominator, then cancel any common factors.

• Conjugates: Use for expressions involving square roots. Multiply the numerator and denominator by the conjugate of the expression containing the root.

• Complex Fractions: Use for fractions within fractions. Find a common denominator to simplify the expression into a single fraction.

• After successfully rewriting the expression and canceling the term causing the indeterminacy, attempt direct substitution again on the simplified expression to find the limit.

Step-by-Step Example 1: Using Factoring

Evaluate the limit: $$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-5x+6}$$

Step 1: Attempt Direct Substitution

Substitute $x=3$ into the expression:

$$\frac{(3{)}^2-9}{(3{)}^2-5(3)+6}=\frac{9-9}{9-15+6}=\frac{0}{0}$$

This is an indeterminate form, so we must use algebraic manipulation.

Step 2: Apply the Algebraic Technique (Factoring)

The numerator is a difference of squares and the denominator is a factorable trinomial. Factor both:

$$x^2-9=(x-3)(x+3)$$

$$x^2-5x+6=(x-3)(x-2)$$

Now, rewrite the limit with the factored forms:

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

Step 3: Cancel the Common Factor

The term $(x-3)$ is present in both the numerator and the denominator. Since the limit approaches $x=3$ but is not equal to $3$, $x-3$ is not zero, and we can safely cancel this term.

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

Step 4: Evaluate the Limit of the Simplified Expression

Now, use direct substitution on the simplified expression:

$$\frac{3+3}{3-2}=\frac{6}{1}=6$$

Therefore, $$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-5x+6}=6$$.

Step-by-Step Example 2: Using the Conjugate Method

Evaluate the limit: $$\underset{h\to 0}{\lim }\frac{\sqrt{4+h}-2}{h}$$

Step 1: Attempt Direct Substitution

Substitute $h=0$ into the expression:

$$\frac{\sqrt{4+0}-2}{0}=\frac{\sqrt{4}-2}{0}=\frac{2-2}{0}=\frac{0}{0}$$

This is an indeterminate form. The presence of a square root suggests using the conjugate method.

Step 2: Multiply by the Conjugate

The conjugate of $$\sqrt{4+h}-2$$ is $$\sqrt{4+h}+2$$. Multiply the numerator and the denominator by this conjugate:

$$\underset{h\to 0}{\lim }\frac{\sqrt{4+h}-2}{h}\cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$$

Step 3: Simplify the Expression

Multiply the numerators using the difference of squares pattern $(a-b)(a+b)=a^2-b^2$:

$$(\sqrt{4+h}{)}^2-(2{)}^2=(4+h)-4=h$$

The expression becomes:

$$\underset{h\to 0}{\lim }\frac{h}{h(\sqrt{4+h}+2)}$$

Step 4: Cancel the Common Factor

The term $h$ is common to the numerator and denominator. Cancel it:

$$\underset{h\to 0}{\lim }\frac{\cancel{h}}{\cancel{h}(\sqrt{4+h}+2)}=\underset{h\to 0}{\lim }\frac{1}{\sqrt{4+h}+2}$$

Step 5: Evaluate the Limit of the Simplified Expression

Use direct substitution on the final expression:

$$\frac{1}{\sqrt{4+0}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{2+2}=\frac{1}{4}$$

Therefore, $$\underset{h\to 0}{\lim }\frac{\sqrt{4+h}-2}{h}=\frac{1}{4}$$.

Using Your Calculator

The problems for this topic are primarily analytical and are often tested on the no-calculator section of the AP Exam. A calculator should not be used to find the answer, as algebraic work is required. However, a graphing calculator can be an excellent tool for verifying your analytical answer.

To check the result of $$\underset{x\to c}{\lim }f(x)$$:

1. Enter the original function, $f(x)$, into Y1 in your calculator's graphing menu.

2. Use the TABLE feature. Set the table to start at $c$ and use a very small step size (e.g., ΔTbl = 0.001`). Examine the `Y1` values for $x-values just above and just below $c$. They should be approaching your calculated limit.

3. Alternatively, graph the function. Use the TRACE feature to move the cursor very close to $x=c$ from both the left and the right. The corresponding $y$-value displayed on the screen should be very close to your answer. You may see a "hole" in the graph at $x=c$ itself, which visually confirms the reason for the initial indeterminate form.

AP Exam Quick Hit

Common Question Types

• Factoring a Rational Function: A straightforward limit problem where both the numerator and denominator are polynomials that share a common root.

  • Example: Evaluate $$\underset{x\to -2}{\lim }\frac{x^2+5x+6}{x+2}$$.

• Simplifying a Complex Fraction: A limit where the main fraction contains smaller fractions, requiring you to find a common denominator to simplify before you can cancel terms.

  • Example: Evaluate $$\underset{x\to 0}{\lim }\frac{\frac{1}{x+5}-\frac{1}{5}}{x}$$.

• Using the Conjugate with Radicals: A limit involving a square root that results in $$\frac{0}{0}$$, requiring multiplication by the conjugate to resolve.

  • Example: Evaluate $$\underset{x\to 1}{\lim }\frac{x-1}{\sqrt{x+3}-2}$$.

Common Mistakes

• Incorrectly Handling $$\frac{0}{0}$$: Mistakenly concluding that a limit resulting in $$\frac{0}{0}$$ is equal to 0, 1, or is automatically undefined, rather than recognizing it as an indeterminate form that requires more work.

• Factoring Errors: Making simple sign errors or other mistakes when factoring polynomials, which prevents the correct term from being canceled.

• Conjugate Multiplication Errors: Forgetting to multiply both the numerator and the denominator by the conjugate, or making an algebraic error when expanding the terms (e.g., incorrectly squaring the radical or distributing a negative sign).

• Stopping Prematurely: Correctly canceling the problematic term but then forgetting to perform the final step of substituting the limit value into the simplified expression.

• "Canceling" Terms Incorrectly: Trying to cancel individual terms in a sum or difference instead of canceling common factors. For example, incorrectly simplifying $$\frac{x^2-4}{x-2}$$ to $$x-2$$ by "canceling" an $x$ and a $-2$ instead of factoring first.