AP Calculus BC: 1.5 Determining Limits Using Algebraic Properties of Limits – Exam Style questions with Answer- FRQ

Question:

The continuous function f is defined on the closed interval −6 ≤ x ≤ 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5, 3). It is known that the point \((3,3-\sqrt{5})\) is on the graph of f. 

(a) If \(\int_{-6}^{5}f(x)dx=7,\) find the value of \(\int_{-6}^{-2}f(x)dx.\) Show the work that leads to your answer
(b) Evaluate \(\int_{3}^{5}(2f'(x)+4)dx.\)
(c) The function g is given by \(g(x)=\int_{-2}^{x}f(t)dt.\) Find the absolute maximum value of g on the interval − 2 ≤ x ≤ 5. Justify your answer.
(d) Find \(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}.\)

▶️Answer/Explanation

Ans:

(a)

\(\int_{-6}^{-2}f(x)dx=\int_{-6}^{5}f(x)dx-\int_{-2}^{5}f(x)dx\)

\(\int_{-6}^{-2}f(x)dx=7-\left [ \frac{1}{2}-\frac{1}{2}-\frac{1}{4}+\frac{9}{4}+\left ( 9-\frac{9\pi }{4} \right ) \right ]\)

(b)

\(\int_{3}^{5}2f'(x)+4dx=\int_{3}^{5}2(f'(x)+2)dx\)

\(=2\int_{3}^{5}f'(x)+2dx\)

\(=2\cdot \left [ f(x)+2x \right ]_{3}^{5}\)

\(=2\cdot \left [ \left [ f(5)+2(5) \right ]-\left [ f(3)+2(3) \right ] \right ]\)

(c)

g'(x) = f(x)

g'(x) = f(x) = 0

x = -1,    x = 1/2,   x = 5

Since f(x) is continuous on -2 ≤ x ≤ 5, then g(x) = \(\int_{-2}^{5}f(t)dt \) is also continuous on -2 ≤ x ≤ 5. Therefore, by the EVT, g(x) has an absolute max. on -2 ≤ x ≤ 5.

Candidates for abs. max : x = -2, -1, 1/2, 5

\(x = -2 : \int_{-2}^{-2}f(t)dt=0\)

\(x = -1 : \int_{-2}^{-1}f(t)dt=\frac{1}{2}\)

\(x = \frac{1}{2} : \int_{-2}^{1/2}f(t)dt=\frac{1}{2}-\frac{1}{2}-\frac{1}{4}=-\frac{1}{4}\)

\(x = 5 : \int_{-2}^{-5}f(t)dt=-\frac{1}{4}+\frac{9}{4}+\left ( 9-\frac{\pi }{4} \right )=2+9-\frac{9\pi }{4}=11-\frac{9\pi }{4}\)

The absolute maximum value of g is \(11-\frac{9\pi }{4}\) and the absolute maximum occurs at x = 5.

(d)

\(\lim_{x\rightarrow 1}10^{x}-3f'(x)=10-3f'(1)=10-3(2)=4\)

\(\lim_{x\rightarrow 1}f(x)-arctan x=1-\frac{\pi }{4}\)

\(arctan(1)=\frac{\pi }{4}\)

\(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}=\frac{4}{1-\pi /4}\)

Question:

Functions f, g, and h are twice-differentiable functions with g(2) = h(2) = 4. The line \(y = 4+\frac{2}{3}(x-2)\) is tangent to both the graph of g at x = 2 and the graph of h at x = 2.
(a) Find h'(2).
(b) Let a be the function given by a(x) = 3x3h(x). Write an expression for a'(x). Find a'(2).
(c) The function h satisfies \(h(x)=\frac{x^{2}-4}{1-(f(x))^{3}}\) for x ≠ 2. It is known that \(\lim_{x\rightarrow 2}h(x)\) can be evaluated using L’Hospital’s Rule. Use \(\lim_{x\rightarrow 2}h(x)\) to find f(2) and f'(2 ). Show the work that leads to your answers.
(d) It is known that g(x) ≤ h(x) for 1 < x < 3. Let k be a function satisfying g(x) ≤ k(x) ≤ h(x) for 1 < x < 3. Is k continuous at x = 2 ? Justify your answer. 

▶️Answer/Explanation

Ans:

(a)

\(y = \frac{2}{3}(x-2)+4\)

point (2, 4)    slope   2/3

h'(2) = 2/3

(b)

a'(x) = h'(x).3x3 + 9x2 h(x)

a'(2) = h'(2) . 3(2)3 + 9(2)2 h(2)

\(a'(2)=\frac{2}{3}\cdot 3\cdot 2^{3}+9\cdot 2^{2}\cdot 4\)

(c)

\(\lim_{x\rightarrow 2}\frac{x^{2}-4}{1-(f(x))^{3}}= \lim_{x\rightarrow 2}\frac{2x}{-3(f(x))^{2}.f'(x)}\)                                        h (2) = 4

                                                                                                                                                                                                                                     \(\therefore \lim_{x\rightarrow 2} h(x)\) must equal 4

\(\therefore \lim_{x\rightarrow 2} x^{2}-4=0\)

\( \lim_{x\rightarrow 2} 1-(f(x))^{3}=0\)  if f(2) = 1                             \(=\frac{2(2)}{-3(1)^{2}\cdot f'(2)}\)

∴ Indeterminate form 0/0                                                           \(=\frac{4}{-3\cdot f'(2)}\)                                 f'(2) must equal -1/3 so that  \(\lim_{x\rightarrow 2} h(x) = 4\)

and L’Hopital’s rule applies                                                        \(=\frac{4}{-3\cdot -1/3}\)

Such that f(2) = 1                                                                              = 4/1

\(\lim_{x\rightarrow 2} h(x) = 4\) such that f(2) = 1 and f'(2) = -1/3

(d)

2 and h given as twice – differentiable ∴  g and h are continuous for 1 < x < 3

g(x) ≤ k(x) ≤ h(x)

\(\lim_{x\rightarrow 2} g(x) = 4\) because  g continuous and g(2) = 4

\(\lim_{x\rightarrow 2} h(x) = 4\) because h continuous and h(2) = 4

By squeeze theorem, \(\lim_{x\rightarrow 2} k(x) = 4\)

k is between or equal to g and h for 1 < x < 3, given g(2) = 4 h (2) , meaning k(2) must equal 4, and \(\lim_{x\rightarrow 2} k(x) = 4\) so k(x) is continuous at x = 2

Question 

Find the limit: \(\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}\).

▶️Answer/Explanation

\(\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}=\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}.\frac{4+\sqrt{16-x}}{4+\sqrt{16-x}}=\lim_{x\rightarrow 0}\frac{x}{x(4+\sqrt{16-x})}=\lim_{x\rightarrow 0}\frac{1}{(4+\sqrt{16-x})}=\frac{1}{8}\)

Question

 (A) Find the limit: \(\lim_{x\rightarrow \frac{\pi }{4}}\frac{x(1-\tan x)}{\cos x-\sin x}\)

(B) Use the result of (a) to derive an approximation of sec x in terms of
x for values of x near \(\frac{\pi }{4}\)

▶️Answer/Explanation

(A)\( \lim_{x\rightarrow \frac{\pi }{4}}\frac{x(1-\tan x)}{\cos x-\sin x}=\lim_{x\rightarrow \frac{\pi }{4}}\frac{x\left ( \frac{\cos x-\sin x}{\cos x} \right )}{\cos x-\sin x}=\lim_{x\rightarrow \frac{\pi }{4}}\frac{x}{\cos x}=\frac{2\pi }{4\sqrt{2}}=\frac{\pi \sqrt{2}}{4x}\).

(B) Near \(x=\frac{\pi }{4}\) , \(x.\sec x=\frac{\pi \sqrt{2}}{4}\).This implies that \(\sec x=\frac{\pi \sqrt{2}}{4x}\)

Question

 (A) Find the limit: \(\lim_{x\rightarrow \frac{\pi }{4}}\frac{x(1-\tan x)}{\cos x-\sin x}\)

(B) Use the result of (a) to derive an approximation of sec x in terms of x for values of x near \(\frac{\pi }{4}\)

▶️Answer/Explanation

(A)\( \lim_{x\rightarrow \frac{\pi }{4}}\frac{x(1-\tan x)}{\cos x-\sin x}=\lim_{x\rightarrow \frac{\pi }{4}}\frac{x\left ( \frac{\cos x-\sin x}{\cos x} \right )}{\cos x-\sin x}=\lim_{x\rightarrow \frac{\pi }{4}}\frac{x}{\cos x}=\frac{2\pi }{4\sqrt{2}}=\frac{\pi \sqrt{2}}{4x}\).

(B) Near \(x=\frac{\pi }{4}\) , \(x.\sec x=\frac{\pi \sqrt{2}}{4}\).This implies that \(\sec x=\frac{\pi \sqrt{2}}{4x}\)

Question 

 Find the limit: \(\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}\).

▶️Answer/Explanation

\(\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}=\lim_{x\rightarrow 0}\frac{4-\sqrt{16-x}}{x}.\frac{4+\sqrt{16-x}}{4+\sqrt{16-x}}=\lim_{x\rightarrow 0}\frac{x}{x(4+\sqrt{16-x})}=\lim_{x\rightarrow 0}\frac{1}{(4+\sqrt{16-x})}=\frac{1}{8}\)

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