BC Calculus
Chapter 2 Multiple Choice Review Questions
- Class: 5H: BC Calculus
- Author: Peter Atlas
- Text: Calculus Finney, Demana, Waits, Kennedy
- \(\displaystyle \displaystyle \lim_{x\to\infty} \frac{5x^3 + 27}{20x^2 + 10x + 9} = \)
- \( \infty\)
- \(\frac{1}{4} \)
- 3
- 0
- 1
Solution
A
- \( \displaystyle \lim_{x \to 0} \frac{\tan{x}}{x}= \)
- 0
- 1
- \(\pi\)
- \( \infty\)
- Nonexistent because left and right-handed limits disagree
Solution
B
- \( \displaystyle \lim_{x \to 0} \frac{\sin{(2x)}}{x} = \)
- 1
- 2
- \(\frac{1}{2} \)
- 0
- \(\infty\)
Solution
B
- \( \displaystyle \lim_{x \to 0} \sin{\frac{1}{x}} = \)
- \( \infty \)
- 1
- nonexistent due to oscillation
- -1
- none of these
Solution
C
- \( \displaystyle \lim_{x \to 0} \tan{ \left( \frac{\pi x}{ x} \right) } = \)
- \( \frac{1}{\pi}\)
- 0
- 1
- \(\pi\)
- \(\infty\)
Solution
B
- Let \(f(x) =
\begin{cases}
x^2 - 1, & \text{if } x \neq 1 \\
4, & \text{if }x = 1
\end{cases}\). Which of the following statements are true?
- \( \displaystyle \lim_{x \to 1} f(x)\) exists.
- \(f(1)\) exists.
- \(f\) is continuous at \(x = 1\).
- only I
- only II
- I and II
- none of them
- all of them
Solution
C
- If \(f(x) =
\begin{cases}
\frac{x^2 - x}{2x}, & \text{if } x \neq 0 \\
k, & \text{if }x = 0
\end{cases}\), then if \(f\) is continuous at \(x = 0\text{, }k = \)
- -1
- \( -\frac{1}{2}\)
- 0
- \(\frac{1}{2} \)
- 1
Solution
B
- Supppose\( \displaystyle f(x) =
\begin{cases}
\frac{3x(x-1)}{x^2 - 3x + 2}, & \text{if } x \neq \{1, 2\} \\
-3, & \text{if } x = 1 \\
4, & \text{if } x = 2
\end{cases}\). Then \(f(x)\) is continous...
- except at \(x = 1\)
- except at \(x = 2\)
- except at \(x = 1\text{ or }2\)
- except at \( x = 0, 1 \text{, or }2\)
- at each real number
Solution
B
- \( \displaystyle \lim_{x \to \infty} \frac {\sin{x}}{x^2 - 3x + 1} = \)
- 0
- 1
- \( \pi \)
- \( \infty \)
Solution
A
- Given \( \displaystyle f(x) \) is defined on \( [-1, 12] \) and that \( f(-1) = 4,\) and \(f(12) = 10, \) which of the following must be true?
- \( \displaystyle \lim_{x \to -1} = 4 \)
- \( \displaystyle \lim_{x \to 0} f(x) = f(0) \)
- \( \displaystyle \left. \exists m \in [-1, 12] \right| f(m) = 6 \)
- The average rate of change of \( f(x) \) on \( [-1, 12] \) is \( \displaystyle \frac {6}{13} \)
Solution
D. Note that the stem of the question never states that \( f \) is continuous.
- Which of the following is a statement of the Sandwich Theorem?
- If \( p(x) \lt k(x) \lt q(x)\) on x \( \in [a, b],\) \( c \in (a, b), \) and \( \displaystyle \lim_{x \to c} p(x) = \lim_{x \to c} q(x) = L, \) then \( \displaystyle \lim_{x \to c} k(x) = L.\)
- If \( p(x) \le k(x) \le q(x)\) on x \( \in [a, b],\) \( c \in [a, b], \) and \( \displaystyle \lim_{x \to c} p(x) = \lim_{x \to c} q(x) = L, \) then \( \displaystyle \lim_{x \to c} k(x) = L.\)
- If \( p(x) \le k(x) \le q(x)\) on x \( \in [a, b],\) then \( \displaystyle \left. \exists c \in (a, b) \right| \lim_{x \to c} k(x) = \lim_{x \to c} p(x) \).
- None of these.
Solution
B. If you got this wrong, look up and memorize the Sandwich Theorem.
- Given \( p(x) \) is any polynomial function, \( \displaystyle \lim_{x \to \infty} \frac {\cos{x}}{p(x)} = \)
- 0
- 1
- \( \pi \)
- Does Not Exist
Solution
A