Calculus BC ยท Unit 1
When substitution gives 0/0, factor, cancel, or rationalize, then substitute again.
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0 over 0 is not the answer
If substituting gives \(\dfrac{0}{0}\), factor, cancel, or rationalize. Then substitute again.
Cancel matching factors
A common factor in the numerator and denominator is 1, as long as \(x \neq a\). Limits do not care about the missing point.
Factor a power of x
If every term has \(x\) or \(x^2\), factor that power out of the top and bottom, then cancel.
Complex fractions
Get a single fraction in the numerator first. Then divide by the leftover \(x\).
Factor a difference of squares, cancel, then substitute.
Factor.
\(\dfrac{x - 3}{(x - 3)(x + 3)}\)
Cancel \(x - 3\).
\(\dfrac{1}{x + 3}\)
Substitute \(x = 3\).
\(\dfrac{1}{3 + 3}\)
\(\dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
Factor.
\(\dfrac{x + 4}{(x - 4)(x + 4)}\)
Cancel \(x + 4\).
\(\dfrac{1}{x - 4}\)
Substitute \(x = -4\).
\(\dfrac{1}{-4 - 4}\)
\(-\dfrac{1}{8}\)
Answer: \(-\dfrac{1}{8}\)
Factor.
\(\dfrac{(x - 5)(x + 5)}{x - 5}\)
Cancel \(x - 5\).
\(x + 5\)
Substitute \(x = 5\).
\(5 + 5\)
\(10\)
Answer: \(10\)
Factor.
\(\dfrac{x(x - 4)}{x}\)
Cancel \(x\).
\(x - 4\)
Substitute \(x = 0\).
\(0 - 4\)
\(-4\)
Answer: \(-4\)
Factor.
\(\dfrac{(x + 2)(x + 3)}{x + 2}\)
Cancel \(x + 2\).
\(x + 3\)
Substitute \(x = -2\).
\(-2 + 3\)
\(1\)
Answer: \(1\)
Factor.
\(\dfrac{2(x - 4)}{(x - 4)(x + 4)}\)
Cancel \(x - 4\).
\(\dfrac{2}{x + 4}\)
Substitute \(x = 4\).
\(\dfrac{2}{8}\)
\(\dfrac{1}{4}\)
Answer: \(\dfrac{1}{4}\)
Cancel a power of x, or factor a difference of cubes.
Factor \(x^2\) from the top and bottom.
\(\dfrac{x^2(4x + 6)}{x^2(2x^2 - 18)}\)
Cancel \(x^2\).
\(\dfrac{4x + 6}{2x^2 - 18}\)
Substitute \(x = 0\).
\(\dfrac{6}{-18}\)
\(-\dfrac{1}{3}\)
Answer: \(-\dfrac{1}{3}\)
Factor \(x^2\).
\(\dfrac{x^2(3x - 12)}{x^2(x^2 - 16)}\)
Cancel \(x^2\).
\(\dfrac{3x - 12}{x^2 - 16}\)
Substitute \(x = 0\).
\(\dfrac{-12}{-16}\)
\(\dfrac{3}{4}\)
Answer: \(\dfrac{3}{4}\)
Factor the difference of cubes.
\(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)
\(\dfrac{x - 2}{(x - 2)(x^2 + 2x + 4)}\)
Cancel \(x - 2\).
\(\dfrac{1}{x^2 + 2x + 4}\)
Substitute \(x = 2\).
\(\dfrac{1}{4 + 4 + 4}\)
\(\dfrac{1}{12}\)
Answer: \(\dfrac{1}{12}\)
\(x^3 + 1 = (x + 1)(x^2 - x + 1)\)
\(\dfrac{x + 1}{(x + 1)(x^2 - x + 1)}\)
Cancel \(x + 1\).
\(\dfrac{1}{x^2 - x + 1}\)
Substitute \(x = -1\).
\(\dfrac{1}{1 + 1 + 1}\)
\(\dfrac{1}{3}\)
Answer: \(\dfrac{1}{3}\)
Factor.
\(\dfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)}\)
Cancel \(x + 3\).
\(\dfrac{x - 3}{x - 2}\)
Substitute \(x = 3\).
\(\dfrac{3 - 3}{3 - 2}\)
\(\dfrac{0}{1}\)
\(0\)
Answer: \(0\)
Factor \(x\).
\(\dfrac{x(2x + 8)}{x^2(x + 4)}\)
\(\dfrac{2(x + 4)}{x(x + 4)}\)
Cancel \(x + 4\).
\(\dfrac{2}{x}\)
Substitute \(x = 0\) makes the denominator 0, and nothing remains to cancel.
The simplified form is \(\dfrac{2}{x}\), which has a vertical asymptote at 0.
does not exist
Answer: does not exist
A complex fraction, or rationalize a square root.
Common denominator in the numerator.
\(\dfrac{1}{4 + x} - \dfrac{1}{4} = \dfrac{4 - (4 + x)}{4(4 + x)}\)
\(\dfrac{-x}{4(4 + x)}\)
Divide by \(x\).
\(\dfrac{-x}{4(4 + x)} \cdot \dfrac{1}{x}\)
\(-\dfrac{1}{4(4 + x)}\)
Substitute \(x = 0\).
\(-\dfrac{1}{4(4)}\)
\(-\dfrac{1}{16}\)
Answer: \(-\dfrac{1}{16}\)
\(\dfrac{1}{5 + x} - \dfrac{1}{5} = \dfrac{5 - (5 + x)}{5(5 + x)}\)
\(\dfrac{-x}{5(5 + x)}\)
\(\dfrac{-x}{5(5 + x)} \cdot \dfrac{1}{x} = -\dfrac{1}{5(5 + x)}\)
Substitute \(x = 0\).
\(-\dfrac{1}{25}\)
Answer: \(-\dfrac{1}{25}\)
Multiply the top and bottom by \(\sqrt{x} + 3\).
\(\dfrac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}\)
\(x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)\)
\(\dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}\)
Cancel \(\sqrt{x} - 3\).
\(\sqrt{x} + 3\)
Substitute \(x = 9\).
\(3 + 3\)
\(6\)
Answer: \(6\)
Multiply the top and bottom by \(\sqrt{x + 9} + 3\).
\(\dfrac{(x + 9) - 9}{x\bigl(\sqrt{x + 9} + 3\bigr)}\)
\(\dfrac{x}{x\bigl(\sqrt{x + 9} + 3\bigr)}\)
Cancel \(x\).
\(\dfrac{1}{\sqrt{x + 9} + 3}\)
Substitute \(x = 0\).
\(\dfrac{1}{3 + 3}\)
\(\dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
\(x^2 - 1 = (x - 1)(x + 1)\)
\(x - 1 = \bigl(\sqrt{x} - 1\bigr)\bigl(\sqrt{x} + 1\bigr)\)
\(\dfrac{\bigl(\sqrt{x} - 1\bigr)\bigl(\sqrt{x} + 1\bigr)(x + 1)}{\sqrt{x} - 1}\)
Cancel \(\sqrt{x} - 1\).
\(\bigl(\sqrt{x} + 1\bigr)(x + 1)\)
Substitute \(x = 1\).
\((1 + 1)(1 + 1)\)
\(2 \cdot 2\)
\(4\)
Answer: \(4\)
\(x^3 - 64 = (x - 4)(x^2 + 4x + 16)\)
\(\dfrac{x - 4}{(x - 4)(x^2 + 4x + 16)}\)
Cancel \(x - 4\).
\(\dfrac{1}{x^2 + 4x + 16}\)
Substitute \(x = 4\).
\(\dfrac{1}{16 + 16 + 16}\)
\(\dfrac{1}{48}\)
Answer: \(\dfrac{1}{48}\)
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