Home / AP Calculus BC 1.8 Determining Limits Using the Squeeze Theorem

AP Calculus BC 1.8 Determining Limits Using the Squeeze Theorem– Exam Style questions with Answer- FRQ

Question • (Hard)

A particle moves along the \( x \)-axis with acceleration \( a(t) = 12t^2 – 4 \) for \( t \geq 0 \). Initial velocity \( v(0) = 0 \), position \( x(1) = 3 \).

(a) Find velocity \( v(t) \).
(b) Given \( -4t \leq v(t) \leq -4t + 4t^3 \) for \( 0 \leq t \leq 0.5 \), use the Squeeze Theorem to find \( \lim_{t \to 0^+} \frac{v(t)}{t} \).
(c) Find position \( x(t) \).

▶️ Answer/Explanation

Solution

(a) \( v(t) = \int a(t) \, dt = \int (12t^2 – 4) \, dt = 4t^3 – 4t + C \). Since \( v(0) = 0 \), \( C = 0 \). \( \boxed{v(t) = 4t^3 – 4t} \).

(b) Inequality: \( -4t \leq v(t) \leq -4t + 4t^3 \). Divide by \( t \): \( -4 \leq \frac{v(t)}{t} \leq -4 + 4t^2 \). Limits as \( t \to 0^+ \): \( -4 \to -4 \), \( -4 + 4t^2 \to -4 \). By Squeeze Theorem: \( \boxed{\lim_{t \to 0^+} \frac{v(t)}{t} = -4} \).

(c) \( x(t) = \int v(t) \, dt = \int (4t^3 – 4t) \, dt = t^4 – 2t^2 + C \). Given \( x(1) = 3 \), \( C = 4 \). \( \boxed{x(t) = t^4 – 2t^2 + 4} \).

Scroll to Top