Factoring Technique for Limits
About This MicroSim
This visualization demonstrates the factoring technique for evaluating limits that produce the indeterminate form 0/0. The example shows:
\[\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\]
The Factoring Process
- Original: \(\frac{x^2 - 9}{x - 3}\) gives 0/0 at x = 3
- Factor: \(\frac{(x+3)(x-3)}{(x-3)}\)
- Cancel: \(x + 3\) (for \(x \neq 3\))
- Substitute: \(\lim_{x \to 3} (x + 3) = 6\)
Visual Insight
- The blue curve with hole shows the original function f(x)
- The green line shows the simplified function g(x) = x + 3
- They're identical except at x = 3!
How to Use
- View Toggle: Switch between original, simplified, or both functions
- X Slider: Move a point along the curve toward x = 3
- Algebra Panel: See the factoring steps
Lesson Plan
Learning Objectives
After using this MicroSim, students will be able to:
- Explain how factoring eliminates the 0/0 indeterminate form
- Understand that factoring doesn't change the limit value
- Apply the factoring technique to difference-of-squares problems
Key Concept
When both numerator and denominator equal zero at x = c, then (x - c) is a factor of both. Canceling this common factor removes the discontinuity and reveals the limit.
Suggested Activities
- Compare Views: Toggle between "Original" and "Simplified" to see how factoring affects the graph
- Approach from Both Sides: Move the slider from left and right of x = 3
- Predict the Limit: Before seeing the algebra, predict what the limit should be based on the graph
Assessment Questions
- Why can we cancel (x - 3) from numerator and denominator?
- Find \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\) using factoring
- What type of discontinuity does the original function have at x = 3?
Embedding
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