Calculus BC ยท Unit 1
Substitution, factoring, a conjugate, a standard trig limit, or the Squeeze Theorem. Match the method to the form.
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Name the form before you calculate
Substitute first. If you get a number, that number is the limit. If you get \(\tfrac{0}{0}\), the expression still needs algebra.
Factor, then cancel
For a rational expression, factor the numerator and the denominator. Cancel the shared factor. Then substitute.
A square root calls for a conjugate
Multiply the numerator and the denominator by the conjugate of the root expression. Then simplify, and then substitute.
Two standard trig limits
\(\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\)
\(\lim_{\theta \to 0} \dfrac{1 - \cos \theta}{\theta^2} = \dfrac{1}{2}\)
Rewrite the problem so one of those blocks appears.
Factor and cancel a shared binomial.
Substitute \(x = 5\): numerator \(0\), denominator \(0\).
\(x^2 - 25 = (x - 5)(x + 5)\)
\(\dfrac{(x - 5)(x + 5)}{x - 5} = x + 5\) for \(x \neq 5\)
Substitute \(x = 5\).
\(5 + 5 = 10\)
Answer: \(10\)
Substitute \(x = -2\): both numerator and denominator are 0.
\(x^2 - 4 = (x - 2)(x + 2)\)
\(\dfrac{(x - 2)(x + 2)}{x + 2} = x - 2\) for \(x \neq -2\)
Substitute \(x = -2\).
\(-2 - 2 = -4\)
Answer: \(-4\)
Substitute \(x = 3\): both numerator and denominator are 0.
\(x^2 - 9 = (x - 3)(x + 3)\)
\(x^2 - 5x + 6 = (x - 2)(x - 3)\)
\(\dfrac{x + 3}{x - 2}\) for \(x \neq 3\)
Substitute \(x = 3\).
\(\dfrac{3 + 3}{3 - 2} = 6\)
Answer: \(6\)
Substitute \(x = 0\): numerator \(0\), denominator \(0\).
Multiply the numerator and the denominator by \(\sqrt{x + 9} + 3\).
\(\dfrac{(x + 9) - 9}{x\bigl(\sqrt{x + 9} + 3\bigr)}\)
\(\dfrac{1}{\sqrt{x + 9} + 3}\) for \(x \neq 0\)
Substitute \(x = 0\).
\(\dfrac{1}{3 + 3} = \dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
For \(x \lt 4\), \(x - 4\) is negative, so \(|x - 4| = -(x - 4)\).
\(\dfrac{|x - 4|}{x - 4} = \dfrac{-(x - 4)}{x - 4} = -1\)
\(\lim_{x \to 4^-} \dfrac{|x - 4|}{x - 4} = -1\)
Answer: \(-1\)
For \(x > 4\), \(|x - 4| = x - 4\).
\(\dfrac{|x - 4|}{x - 4} = 1\)
\(\lim_{x \to 4^+} \dfrac{|x - 4|}{x - 4} = 1\)
Answer: \(1\)
Pick the algebra that clears the indeterminate form.
Substitute \(x = 2\): numerator \(0\), denominator \(0\).
Common denominator in the numerator:
\(\dfrac{1}{x} - \dfrac{1}{2} = \dfrac{2 - x}{2x}\)
\(\dfrac{\tfrac{2 - x}{2x}}{x - 2} = \dfrac{2 - x}{2x(x - 2)}\)
\(2 - x = -(x - 2)\)
\(\dfrac{-(x - 2)}{2x(x - 2)} = -\dfrac{1}{2x}\) for \(x \neq 2\)
Substitute \(x = 2\).
\(-\dfrac{1}{2 \cdot 2} = -\dfrac{1}{4}\)
Answer: \(-\dfrac{1}{4}\)
Use \(\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\).
\(\dfrac{\sin(5x)}{x} = 5 \cdot \dfrac{\sin(5x)}{5x}\)
As \(x \to 0\), the input \(5x\) also approaches 0.
\(5 \cdot 1 = 5\)
Answer: \(5\)
Substitute \(x = 9\): numerator \(0\), denominator \(0\).
Multiply by the conjugate \(\sqrt{x} + 3\).
\(\dfrac{x - 9}{(x - 9)\bigl(\sqrt{x} + 3\bigr)} = \dfrac{1}{\sqrt{x} + 3}\) for \(x \neq 9\)
Substitute \(x = 9\).
\(\dfrac{1}{3 + 3} = \dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
Substitute \(x = -3\): both numerator and denominator are 0.
\(x^2 + x - 6 = (x + 3)(x - 2)\)
\(x - 2\) for \(x \neq -3\)
Substitute \(x = -3\).
\(-3 - 2 = -5\)
Answer: \(-5\)
Use the identity \(1 - \cos x = 2\sin^2\!\bigl(\tfrac{x}{2}\bigr)\).
\(\dfrac{1 - \cos x}{x^2} = \dfrac{2\sin^2\!\bigl(\tfrac{x}{2}\bigr)}{x^2}\)
\(\dfrac{2\sin^2\!\bigl(\tfrac{x}{2}\bigr)}{4\bigl(\tfrac{x}{2}\bigr)^2} = \dfrac{1}{2}\left(\dfrac{\sin\!\bigl(\tfrac{x}{2}\bigr)}{\tfrac{x}{2}}\right)^2\)
As \(x \to 0\), \(\tfrac{x}{2} \to 0\), so the sine ratio approaches 1.
\(\dfrac{1}{2} \cdot 1^2 = \dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
Direct substitution works when the expression is defined at that x-value.
A gives \(\tfrac{0}{0}\).
B gives \(\tfrac{0}{0}\).
C gives \(\dfrac{9 + 4}{3 - 1} = \dfrac{13}{2}\).
D gives \(\tfrac{0}{0}\).
Answer: C. \(\dfrac{13}{2}\)
Absolute value, a squeeze, a complex fraction, or a quotient of sines.
\(\lim_{x \to 4^-} \dfrac{|x - 4|}{x - 4} = -1\)
\(\lim_{x \to 4^+} \dfrac{|x - 4|}{x - 4} = 1\)
The one-sided limits are not equal.
The two-sided limit does not exist.
Answer: does not exist
\(-1 \leq \sin\!\bigl(\tfrac{1}{x}\bigr) \leq 1\)
Multiply through by \(4x^2\), which is nonnegative.
\(-4x^2 \leq 4x^2\sin\!\bigl(\tfrac{1}{x}\bigr) \leq 4x^2\)
\(\lim_{x \to 0} (-4x^2) = 0\)
\(\lim_{x \to 0} (4x^2) = 0\)
The bounds have the same limit, so the Squeeze Theorem applies.
\(0\)
Answer: \(0\)
Substitute \(x = 0\): numerator \(0\), denominator \(0\).
Multiply by \(\sqrt{x + 4} + 2\).
\(\dfrac{1}{\sqrt{x + 4} + 2}\) for \(x \neq 0\)
Substitute \(x = 0\).
\(\dfrac{1}{2 + 2} = \dfrac{1}{4}\)
Answer: \(\dfrac{1}{4}\)
Substitute \(x = 3\): both fractions are undefined.
\(x^2 - 9 = (x - 3)(x + 3)\)
Common denominator \((x - 3)(x + 3)\):
\(\dfrac{x + 3}{(x - 3)(x + 3)} - \dfrac{6}{(x - 3)(x + 3)}\)
\(\dfrac{x + 3 - 6}{(x - 3)(x + 3)} = \dfrac{x - 3}{(x - 3)(x + 3)}\)
\(\dfrac{1}{x + 3}\) for \(x \neq 3\)
Substitute \(x = 3\).
\(\dfrac{1}{3 + 3} = \dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
Use \(\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\).
\(\dfrac{\sin(4x)}{\sin(7x)} = \dfrac{4}{7} \cdot \dfrac{\sin(4x)}{4x} \cdot \dfrac{7x}{\sin(7x)}\)
As \(x \to 0\), both \(4x\) and \(7x\) approach 0.
\(\dfrac{4}{7} \cdot 1 \cdot 1 = \dfrac{4}{7}\)
Answer: \(\dfrac{4}{7}\)
Substitute \(x = 2\).
\(\dfrac{4 + 1}{2 - 5} = \dfrac{5}{-3} = -\dfrac{5}{3}\)
The expression is defined at \(x = 2\), so substitution is the whole method.
Answer: C. \(-\dfrac{5}{3}\)
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