AcademyLimits
Academy
Algebraic Limits
Level 1 - Math I (Physics) topic page in Limits.
Algebraic Methods for Evaluating Limits
While direct substitution works for many functions, algebraic manipulation is often necessary when direct substitution produces an indeterminate form like \(\frac{0}{0}\).
Direct Substitution
For polynomial functions, we can substitute the value directly:
Direct Substitution
\[\lim_{x \to 2} (3x^2 - 5x + 1) = 3(2)^2 - 5(2) + 1 = 12 - 10 + 1 = 3\]
For rational functions, direct substitution works when the denominator is non-zero:
Rational Limit
\[\lim_{x \to 3} \frac{x^2 + 1}{x - 1} = \frac{3^2 + 1}{3 - 1} = \frac{10}{2} = 5\]
Factoring Method
When direct substitution gives \(\frac{0}{0}\), factor and cancel:
Factoring Example
\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4\]
The key is identifying the common factor that causes the zero in the denominator, then canceling it before taking the limit.
Rationalizing
For expressions with square roots, rationalize by multiplying by the conjugate:
Rationalizing
\[\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \lim_{x \to 4} \frac{x - 4}{(x-4)(\sqrt{x} + 2)} = \lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{4}\]
Difference of Cubes and Special Products
For expressions involving differences of cubes:
Difference of Cubes
\[\lim_{x \to 8} \frac{x^{2/3} - 4}{x - 8} = \lim_{x \to 8} \frac{(x^{1/3} - 2)(x^{1/3} + 2)}{x - 8}\]
Let \(u = x^{1/3}\), so when \(x \to 8\), \(u \to 2\):
Substituted Form
\[= \lim_{u \to 2} \frac{(u-2)(u+2)}{u^3 - 8} = \lim_{u \to 2} \frac{u+2}{u^2 + 2u + 4} = \frac{4}{12} = \frac{1}{3}\]
Strategy Summary
- Try direct substitution first
- If result is \(\frac{0}{0}\), factor and cancel
- If radicals are present, rationalize
- Consider substitution for complex expressions
- Apply limit laws after simplification