Lesson PreviewLesson 6 of 16 in this unit

1.6 Determining Limits Using Algebraic Manipulation

Part of Limits and Continuity in AP® Calculus AB. Read the opening section, then create a free account for the full interactive lesson, practice, and progress tracking.

Start free

Sometimes direct substitution gives the indeterminate form

\frac{0}{0}

This does not mean the limit does not exist. It means the expression must be rewritten first. In this lesson, the main algebraic tools are factoring, canceling common factors, combining fractions, and rationalizing with conjugates. The goal is to simplify the expression so the actual behavior near the point becomes visible.

This works because a limit depends on values near the point, not necessarily at the point itself. After simplifying, the new expression may not be defined exactly the same way at the original point, but that is okay for limits. What matters is that the simplified expression matches the original one for nearby x-values, allowing us to determine the approach behavior correctly.

Example 1: Find

\lim_{x \to 3}\frac{x^2-9}{x-3}

Factor the numerator:

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

\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3 \quad \text{for } x\ne3

Now take the limit:

\lim_{x \to 3}(x+3)=6

Example 2: Find

\lim_{x \to 0}\frac{\sqrt{x+1}-1}{x}

Rationalize the numerator:

\frac{\sqrt{x+1}-1}{x}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{(x+1)-1}{x(\sqrt{x+1}+1)}=\frac{x}{x(\sqrt{x+1}+1)}=\frac{1}{\sqrt{x+1}+1}

Now evaluate:

\lim_{x \to 0}\frac{1}{\sqrt{x+1}+1}=\frac{1}{2}

In this unit

16 lessons

Continue the full lesson in Stellar

Create a free account to unlock the remaining sections, checkpoints, course practice, and progress tracking for AP® Calculus AB.

Start free