Home / AP Calculus AB 1.6 Determining Limits Using Algebraic Manipulation – MCQs

AP Calculus AB 1.6 Determining Limits Using Algebraic Manipulation - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

\[ \lim_{x\to 2}\frac{(x+2)(x-2)}{4(x-5)(2-x)} \text{ is} \]
(A) \(-\tfrac{1}{3}\)
(B) \(-\tfrac{1}{4}\)
(C) \(\tfrac{1}{3}\)
(D) nonexistent
▶️ Answer/Explanation
Note \(2-x=-(x-2)\). Then \[ \frac{(x+2)(x-2)}{4(x-5)(2-x)} =\frac{(x+2)(x-2)}{-4(x-5)(x-2)} =\frac{x+2}{-4(x-5)}. \] Evaluate at \(x=2\): \(\dfrac{4}{-4(-3)}=\dfrac{4}{12}=\dfrac{1}{3}\).
✅ Answer: (C)

No-Calc Question

\(\displaystyle \lim_{x\to 0}\frac{2x^{6}+6x^{3}}{4x^{5}+3x^{3}}=\)

(A) \(0\)
(B) \(\tfrac{1}{2}\)
(C) \(1\)
(D) \(2\)
(E) nonexistent

▶️ Answer/Explanation

Factor out \(x^{3}\) from numerator and denominator:
\(\displaystyle \frac{2x^{6}+6x^{3}}{4x^{5}+3x^{3}} =\frac{x^{3}\,(2x^{3}+6)}{x^{3}\,(4x^{2}+3)} =\frac{2x^{3}+6}{4x^{2}+3}\).
Now take \(x\to 0\): \(\displaystyle \frac{2\cdot 0+6}{4\cdot 0+3}=\frac{6}{3}=2\).
✅ Answer: (D)

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