Questions: 23. For x ≥ 0, the horizontal line y=2 is an asymptote for the graph of the function f. Which of the following statements must be true? a. f(0)=2 b. f(x) ≠ 2 for all x ≥ 0 c. f(2) is undefined d. lim x → 2 f(x)=∞ e. lim x → ∞ f(x)=2 24. lim n → ∞ (3n^3-5n)/(n^3-2n^2+1) is a. -5 b. -2 c. 1 d. 3 e. nonexistent 25. lim h → 0 (1/h) ln ((2+h)/2) is a. e^2 b. 1 c. 1/2 d. 0 e. nonexistent

23. For x ≥ 0, the horizontal line y=2 is an asymptote for the graph of the function f. Which of the following statements must be true?
a. f(0)=2
b. f(x) ≠ 2 for all x ≥ 0
c. f(2) is undefined
d. lim x → 2 f(x)=∞
e. lim x → ∞ f(x)=2
24. lim n → ∞ (3n^3-5n)/(n^3-2n^2+1) is
a. -5
b. -2
c. 1
d. 3
e. nonexistent
25. lim h → 0 (1/h) ln ((2+h)/2) is
a. e^2
b. 1
c. 1/2
d. 0
e. nonexistent

Transcript text: 23. For $x \geq 0$, the horizontal line $y=2$ is an asymptote for the graph of the function $f$ Which of the following statements must be true? a. $\quad f(0)=2$ b. $\quad f(x) \neq 2$ for all $x \geq 0$ c. $f(2)$ is undefined d. $\lim _{x \rightarrow 2} f(x)=\infty$ e. $\lim _{x \rightarrow \infty} f(x)=2$ 24. $\lim _{n \rightarrow \infty} \frac{3 n^{3}-5 n}{n^{3}-2 n^{2}+1}$ is a. -5 b. -2 c. 1 d. 3 e. nonexistent 25. $\lim _{h \rightarrow 0} \frac{1}{h} \ln \left(\frac{2+h}{2}\right)$ is a. $e^{2}$ b. 1 c. $\frac{1}{2}$ d. 0 e. nonexistent

Solution

Solution Steps

For $ x \geq 0 $, the horizontal line $ y=2 $ is an asymptote for the graph of the function $ f $. Which of the following statements must be true? e. $ \lim_{x \rightarrow \infty} f(x) = 2 $
To find $ \lim_{n \rightarrow \infty} \frac{3n^3 - 5n}{n^3 - 2n^2 + 1} $, we divide the numerator and the denominator by $ n^3 $ and then take the limit as $ n $ approaches infinity.
To find $ \lim_{h \rightarrow 0} \frac{1}{h} \ln \left(\frac{2+h}{2}\right) $, we can use L'Hôpital's rule since the limit is in the indeterminate form $ \frac{0}{0} $.

Step 1: Analyze the Asymptote Condition

Given that the horizontal line $ y = 2 $ is an asymptote for the graph of the function $ f $ for $ x \geq 0 $, we need to determine which statement must be true.

Step 2: Evaluate the Statements

Statement (a): $ f(0) = 2 $ is not necessarily true because the function can approach the asymptote without touching it at $ x = 0 $.
Statement (b): $ f(x) \neq 2 $ for all $ x \geq 0 $ is not necessarily true because the function can cross the asymptote.
Statement (c): $ f(2) $ is undefined is not necessarily true because the function can be defined at $ x = 2 $.
Statement (d): $ \lim_{x \rightarrow 2} f(x) = \infty $ is not necessarily true because the asymptote is horizontal, not vertical.
Statement (e): $ \lim_{x \rightarrow \infty} f(x) = 2 $ must be true because the function approaches the horizontal asymptote as $ x $ approaches infinity.

Step 3: Compute the Limit for Question 24

To find $ \lim_{n \rightarrow \infty} \frac{3n^3 - 5n}{n^3 - 2n^2 + 1} $, we divide the numerator and the denominator by $ n^3 $:

\[ \lim_{n \rightarrow \infty} \frac{3n^3 - 5n}{n^3 - 2n^2 + 1} = \lim_{n \rightarrow \infty} \frac{3 - \frac{5}{n^2}}{1 - \frac{2}{n} + \frac{1}{n^3}} = \frac{3}{1} = 3 \]

Step 4: Compute the Limit for Question 25

To find $ \lim_{h \rightarrow 0} \frac{1}{h} \ln \left(\frac{2+h}{2}\right) $, we use L'Hôpital's rule:

\[ \lim_{h \rightarrow 0} \frac{1}{h} \ln \left(\frac{2+h}{2}\right) = \lim_{h \rightarrow 0} \frac{\ln \left(\frac{2+h}{2}\right)}{h} \]

Taking the derivative of the numerator and the denominator:

\[ \lim_{h \rightarrow 0} \frac{\frac{d}{dh} \ln \left(\frac{2+h}{2}\right)}{\frac{d}{dh} h} = \lim_{h \rightarrow 0} \frac{\frac{1}{\frac{2+h}{2}} \cdot \frac{1}{2}}{1} = \lim_{h \rightarrow 0} \frac{1}{2+h} = \frac{1}{2} \]