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

2. $\lim_{x \to -2} \frac{3x^2 + 5x + 7}{x - 4}$ is
a) $\frac{-5}{2}$
b) $\frac{-3}{2}$
c) -1
d) 3

3. 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

4. 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

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

2. $\lim_{x \to -2} \frac{3x^2 + 5x + 7}{x - 4}$ is
   a) $\frac{-5}{2}$
   b) $\frac{-3}{2}$
   c) -1
   d) 3

3. 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

4. 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

Accepted Answer

### Question 1
# Explanation:
## 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 $
# Answer:
a) 0

### Question 2
# Explanation:
## 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} $
# Answer:
b) $ -\frac{3}{2} $

### Question 3
# Explanation:
## 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:
d) I, II, and III

### Question 4
# Explanation:
## Step1: Evaluate $ \lim_{x \to 1}\frac{x^{2}+1}{x - 1} $
As $ x\to1 $, the denominator $ x - 1\to0 $ and the numerator $ x^{2}+1\to1^{2}+1 = 2
eq0 $. When the denominator approaches 0 and the numerator approaches a non - zero number, the limit $ \lim_{x \to 1}\frac{x^{2}+1}{x - 1} $ does not exist (it approaches $ \pm\infty $)
## Step2: Evaluate $ \lim_{x \to 0}\frac{|x|}{x} $
We consider the left - hand limit and the right - hand limit.
- For $ x\to0^{+} $ ( $ x>0 $ ), $ |x|=x $, so $ \frac{|x|}{x}=\frac{x}{x}=1 $, $ \lim_{x \to 0^{+}}\frac{|x|}{x}=1 $
- For $ x\to0^{-} $ ( $ x<0 $ ), $ |x|=-x $, so $ \frac{|x|}{x}=\frac{-x}{x}=-1 $
Since the left - hand limit ($ - 1$) and the right - hand limit ($ 1$) are not equal, $ \lim_{x \to 0}\frac{|x|}{x} $ does not exist.
## Step3: Evaluate $ \lim_{x \to 2}f(x) $ where $ f(x)=\begin{cases}3&x\leq2\0&x > 2\end{cases} $
- Left - hand limit: $ \lim_{x \to 2^{-}}f(x)=3 $ (since for $ x\leq2 $, $ f(x) = 3$)
- Right - hand limit: $ \lim_{x \to 2^{+}}f(x)=0 $ (since for $ x>2 $, $ f(x)=0$)
Since $ 3
eq0 $, $ \lim_{x \to 2}f(x) $ does not exist.

All three limits do not exist.
# Answer:
d) I, II, and III

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 1

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

Step2: Evaluate \( \sin \pi \)

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

Step2: Simplify the fraction

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

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

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

Answer:

Question 2