AP Precalculus • Vertical Asymptotes & One-Sided Limit Behavior • Solutions
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Problem 4 Solution \( r(x) = \dfrac{x+2}{x-5} \) at \( x = 5 \)
a) \( \displaystyle \lim_{x \to 5^-} r(x) \):
Test \( x = 4.99 \):
• Numerator: \( 4.99 + 2 = 6.99 > 0 \) (Positive)
• Denominator: \( 4.99 - 5 = -0.01 < 0 \) (Small Negative, \( 0^- \))
\( \dfrac{(+)}{(0^-)} \implies \lim_{x \to 5^-} r(x) = -\infty \)
b) \( \displaystyle \lim_{x \to 5^+} r(x) \):
Test \( x = 5.01 \):
• Numerator: \( 5.01 + 2 = 7.01 > 0 \) (Positive)
• Denominator: \( 5.01 - 5 = +0.01 > 0 \) (Small Positive, \( 0^+ \))
\( \dfrac{(+)}{(0^+)} \implies \lim_{x \to 5^+} r(x) = +\infty \)
Problem 5 Solution \( k(x) = \dfrac{3-x}{(x+2)^2} \)
a) Left Limit \( x \to -2^- \):
Numerator: \( 3 - (-2) = 5 > 0 \)
Denominator: \( (0^-)^2 \to 0^+ \)
\( \lim_{x \to -2^-} k(x) = +\infty \)
b) Right Limit \( x \to -2^+ \):
Numerator: \( 3 - (-2) = 5 > 0 \)
Denominator: \( (0^+)^2 \to 0^+ \)
\( \lim_{x \to -2^+} k(x) = +\infty \)
c) Visual Behavior:
The branches point in the same direction (both \( +\infty \)).
Reason: Even exponent (power of 2) prevents denominator from changing sign across \( x = -2 \).
Problem 6 Solution \( R(x) = \dfrac{x-1}{(x-3)(x+2)} \)
a) \( x \to -2^- \):
Sign: \( \frac{(-)}{(-)(-)} = \frac{(-)}{(+)} \)
\( -\infty \)
b) \( x \to -2^+ \):
Sign: \( \frac{(-)}{(-)(+)} = \frac{(-)}{(-)} \)
\( +\infty \)
c) \( x \to 3^- \):
Sign: \( \frac{(+)}{(-)(+)} = \frac{(+)}{(-)} \)
\( -\infty \)
d) \( x \to 3^+ \):
Sign: \( \frac{(+)}{(+)(+)} = \frac{(+)}{(+)} \)
\( +\infty \)
Problem 7 Solution \( p(x) = \dfrac{x(x-1)(x+1)}{x(x-1)(x-2)} = \dfrac{x+1}{x-2} \) (\(x \neq 0, 1\))
a) Classify Discontinuities:
• \( x = 0 \): Removable (Hole)
• \( x = 1 \): Removable (Hole)
• \( x = 2 \): Vertical Asymptote
b) Exact Hole Coordinates:
At \( x = 0 \): \( \frac{0+1}{0-2} = -\frac{1}{2} \implies \mathbf{\left(0, -\frac{1}{2}\right)} \)
At \( x = 1 \): \( \frac{1+1}{1-2} = -2 \implies \mathbf{(1, -2)} \)
c) Limits at VA \( x = 2 \):
• \( \displaystyle \lim_{x \to 2^-} p(x) = \frac{3}{0^-} = \mathbf{-\infty} \)
• \( \displaystyle \lim_{x \to 2^+} p(x) = \frac{3}{0^+} = \mathbf{+\infty} \)
AP Precalculus • Topic 1.9 & 1.10 Answer Key Page 2 of 3
AP Precalculus • Advanced Limit Behavior & Function Synthesis • Solutions
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Problem 8 Solution \( q(x) = \dfrac{2x^3 - 8x}{x^3 + x^2 - 6x} = \dfrac{2x(x-2)(x+2)}{x(x+3)(x-2)} \)
a) Factoring & Domain:
Simplified: \( \tilde{q}(x) = \dfrac{2(x+2)}{x+3} \) for \( x \neq 0, 2 \)
Domain: \( \{x \mid x \neq -3, 0, 2\} \)
b) Coordinates of Holes:
\( x = 0 \): \( y = \frac{2(2)}{3} = \frac{4}{3} \implies \mathbf{\left(0, \frac{4}{3}\right)} \)
\( x = 2 \): \( y = \frac{2(4)}{5} = \frac{8}{5} \implies \mathbf{\left(2, \frac{8}{5}\right)} \)
c) \( \displaystyle \lim_{x \to 0} q(x) \) \( \mathbf{\dfrac{4}{3}} \)
d) \( \displaystyle \lim_{x \to -3^-} q(x) \)
\( \frac{2(-1)}{0^-} = \frac{-2}{0^-} \)
\( \mathbf{+\infty} \)
e) \( \displaystyle \lim_{x \to -3^+} q(x) \)
\( \frac{2(-1)}{0^+} = \frac{-2}{0^+} \)
\( \mathbf{-\infty} \)
Problem 9 Solution Function Synthesis
Step 1: Set Up Factors from Constraints
Form: \( M(x) = \dfrac{a(x+1)(x-3)}{(x+4)(x-3)} \)
Step 2: Solve for Leading Coefficient \( a \)
The hole has coordinates \( (3, 2) \). In simplified form:
\( \lim_{x \to 3} M(x) = \dfrac{a(3+1)}{3+4} = \dfrac{4a}{7} = 2 \)
\( 4a = 14 \implies a = \dfrac{7}{2} = 3.5 \)
Check VA sign: as \( x \to -4^- \), \( \frac{3.5(-3.1)}{-0.1} > 0 \implies +\infty \). Confirmed!
Final Answer: \( M(x) = \dfrac{7(x+1)(x-3)}{2(x+4)(x-3)} = \dfrac{3.5(x+1)(x-3)}{(x+4)(x-3)} \)
Problem 10 Solution AP CED Style • Limit Table Reasoning
Given \( W(x) = \dfrac{(x-a)(x+b)}{(x-a)(x-c)^2} \):
a) Values of \( a \) and \( c \):
Near \( x = 1 \), \( W(x) \to 0.750 \) (finite limit) → hole at \( x = 1 \):
\( a = 1 \)
Near \( x = -1 \), \( W(x) \to +\infty \) → VA with even power:
\( c = -1 \)
b) Solve for \( b \):
Simplified: \( \tilde{W}(x) = \dfrac{x+b}{(x+1)^2} \)
\( \lim_{x \to 1} W(x) = \dfrac{1+b}{(1+1)^2} = \dfrac{3}{4} \)
\( \dfrac{1+b}{4} = \dfrac{3}{4} \implies 1+b = 3 \)
\( b = 2 \)
c) Asymptote Limits:
From the table and analysis:
\( \displaystyle \lim_{x \to -1^-} W(x) = +\infty \)
\( \displaystyle \lim_{x \to -1^+} W(x) = +\infty \)
Both branches tend to \( +\infty \) due to squared denominator factor.
AP Precalculus • Topic 1.9 & 1.10 Answer Key Page 3 of 3
