AP Calculus AB 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms - MCQs - Exam Style Questions
No-Calc Question
\(\displaystyle \lim_{x\to 0}\frac{e^{2x}-\ln(x+1)-\cos(\pi x)}{\sin(\pi x)}=\)
(A) \(0\)
(B) \(\tfrac{1}{\pi}\)
(C) \(\tfrac{2}{\pi}\)
(D) \(1\)
(B) \(\tfrac{1}{\pi}\)
(C) \(\tfrac{2}{\pi}\)
(D) \(1\)
▶️ Answer/Explanation
It is \(0/0\). Apply L’Hôpital’s Rule.
Derivative (top): \(2e^{2x}-\dfrac{1}{x+1}+\pi\sin(\pi x)\).
Derivative (bottom): \(\pi\cos(\pi x)\).
Evaluate at \(x=0\): \(\dfrac{2-1+0}{\pi\cdot 1}=\dfrac{1}{\pi}\).
✅ Answer: (B)
Derivative (top): \(2e^{2x}-\dfrac{1}{x+1}+\pi\sin(\pi x)\).
Derivative (bottom): \(\pi\cos(\pi x)\).
Evaluate at \(x=0\): \(\dfrac{2-1+0}{\pi\cdot 1}=\dfrac{1}{\pi}\).
✅ Answer: (B)
No-Calc Question
\(\displaystyle \lim_{x\to 0}\frac{\sin x}{e^{x}-1}\) is
(A) \(1\)
(B) \(\tfrac{1}{e}\)
(C) \(0\)
(D) nonexistent
(B) \(\tfrac{1}{e}\)
(C) \(0\)
(D) nonexistent
▶️ Answer/Explanation
Use series/derivatives at \(0\).
\(\sin x\sim x\), \(e^{x}-1\sim x\).
Ratio \(\displaystyle \lim_{x\to0}\frac{x}{x}=1\).
✅ Answer: (A)