Question

chapter 1 ap exam practice
section 1: part a. multiple choice. no calculator

1. $\lim_{x \to \pi} \frac{\sin x}{x}$ is

a) 0
b) 1
c) $\pi$
d) nonexistent

1. $\lim_{x \to -2} \frac{3x^2 + 5x + 7}{x - 4}$ is

a) $\frac{-5}{2}$
b) $\frac{-3}{2}$
c) -1
d) 3

1. consider the function $f(x) = \

$$\begin{cases} \\sqrt{x - 2} & \\text{if } x > 3 \\\\ 6 - 2x & \\text{if } x \\leq 3 \\end{cases}$$

$ which of the following is true?
i. $\lim_{x \to 3} \sqrt{x - 2} = 1$
ii. $\lim_{x \to 3} (6 - 2x) = 0$
iii. the limit $\lim_{x \to 3} f(x)$ does not exist
a) none
b) i and ii only
c) iii only
d) i, ii, and iii

1. which of the following limits do not exist?

i. $\lim_{x \to 1} \frac{x^2 + 1}{x - 1}$
ii. $\lim_{x \to 0} \frac{|x|}{x}$
iii. $\lim_{x \to 2} f(x)$, where $f(x) = \

$$\begin{cases} 3 & \\text{if } x \\leq 2 \\\\ 0 & \\text{if } x > 2 \\end{cases}$$

$
a) none
b) i and ii only
c) iii only
d) i, ii, and iii

Explanation:

Question 1

Step1: Substitute \( x = \pi \) into the function

We have the limit \( \lim_{x \to \pi} \frac{\sin x}{x} \). Since \( \sin x \) and \( x \) are continuous functions (for \( x
eq0 \)), we can use the direct substitution property of limits. Substitute \( x = \pi \) into \( \frac{\sin x}{x} \):
\( \frac{\sin \pi}{\pi} \)

Step2: Evaluate \( \sin \pi \)

We know that \( \sin \pi = 0 \), so the expression becomes \( \frac{0}{\pi}=0 \)

Step1: Substitute \( x=-2 \) into the function

We have the limit \( \lim_{x \to -2} \frac{3x^{2}+5x + 7}{x - 4} \). Since the denominator \( x - 4 \) is not zero when \( x=-2 \) (because \( -2-4=-6
eq0 \)) and the numerator is a polynomial (hence continuous), we can use direct substitution. Substitute \( x = -2 \) into the numerator and the denominator:
Numerator: \( 3(-2)^{2}+5(-2)+7=3\times4-10 + 7=12-10 + 7=9 \)
Denominator: \( -2-4=-6 \)

Step2: Simplify the fraction

The limit is \( \frac{9}{-6}=-\frac{3}{2} \)

Step1: Evaluate \( \lim_{x \to 3^{+}} \sqrt{x - 2} \)

For the right - hand limit (as \( x\to3^{+} \), \( x>3 \)), we use the function \( \sqrt{x - 2} \). Substitute \( x = 3 \) into \( \sqrt{x - 2} \): \( \sqrt{3 - 2}=\sqrt{1}=1 \), so \( \lim_{x \to 3^{+}} \sqrt{x - 2}=1 \) (Statement I is true)

Step2: Evaluate \( \lim_{x \to 3^{-}} (6 - 2x) \)

For the left - hand limit (as \( x\to3^{-} \), \( x\leq3 \)), we use the function \( 6 - 2x \). Substitute \( x = 3 \) into \( 6 - 2x \): \( 6-2\times3=6 - 6 = 0 \), so \( \lim_{x \to 3^{-}}(6 - 2x)=0 \) (Statement II is true)

Step3: Evaluate \( \lim_{x \to 3}f(x) \)

For the limit of a piece - wise function to exist at \( x = 3 \), the left - hand limit and the right - hand limit must be equal. We found that \( \lim_{x \to 3^{+}}f(x)=1 \) and \( \lim_{x \to 3^{-}}f(x)=0 \). Since \( 1
eq0 \), \( \lim_{x \to 3}f(x) \) does not exist (Statement III is true)

Answer:

a) 0

Question 2