AP Calculus BC 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms - Exam Style Questions - MCQs - New Syllabus
Question
\[ \lim_{x\to \pi/4}\ \frac{\displaystyle\int_{\pi/4}^{x}\tan(t)\,dt}{x^{2}-\pi^{2}/16}=\,? \]
(A) \(0\)
(B) \(\tfrac{2}{\pi}\)
(C) \(\tfrac{4}{\pi}\)
(D) nonexistent
▶️ Answer/Explanation
Detailed solution
\(0/0\) form → L’Hôpital.\(\dfrac{d}{dx}\) numerator \(=\tan x\) (FTC).
\(\dfrac{d}{dx}\) denominator \(=2x\).
\[ \lim_{x\to \pi/4}\frac{\tan x}{2x} =\frac{1}{2(\pi/4)}=\frac{2}{\pi}. \] ✅ Answer: (B)
Question
Let \(f\) and \(g\) be twice-differentiable functions such that \(\displaystyle \lim_{x\to\infty} f(x)=\infty,\quad \lim_{x\to\infty} g(x)=\infty,\quad \lim_{x\to\infty} f'(x)=8,\quad \lim_{x\to\infty} g'(x)=2.\)
What is \(\displaystyle \lim_{x\to\infty}\!\left(\frac{f(x)}{g(x)}+6\right)\)?
What is \(\displaystyle \lim_{x\to\infty}\!\left(\frac{f(x)}{g(x)}+6\right)\)?
(A) \(4\)
(B) \(7\)
(C) \(10\)
(D) nonexistent
▶️ Answer/Explanation
Correct Answer: C
Because \(\displaystyle \frac{f(x)}{g(x)}\) is an \(\frac{\infty}{\infty}\) form, apply L’Hôpital’s Rule: \(\displaystyle \lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\frac{8}{2}=4.\)
Therefore \(\displaystyle \lim_{x\to\infty}\!\left(\frac{f(x)}{g(x)}+6\right)=4+6=10.\)