AP Calculus AB 1.6 Determining Limits Using Algebraic Manipulation - MCQs - Exam Style Questions
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
(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)
✅ Answer: (C)