▶️ Answer/Explanation
Solution
Since \( \lim_{x \to 3} f(x) = 6 \), the function values must approach 6 as \( x \) approaches 3 from both sides. Graph C shows the function approaching 6 near \( x = 3 \), matching the limit condition.
Answer: C
▶️ Answer/Explanation
Solution
The limit \( \lim_{x \to 0} f(x) \) is found using L’Hôpital’s Rule: \( \frac{d}{dx}(e^{3x} – 1) = 3e^{3x} \) and \( \frac{d}{dx}(x) = 1 \), so \( \lim_{x \to 0} \frac{3e^{3x}}{1} = 3 \). Thus, \( f(x) \) approaches 3 as \( x \to 0 \), expressed by \( \lim_{x \to 0} f(x) = 3 \).
Answer: C
▶️ Answer/Explanation
Solution
The property “as \( x \) gets closer to 3, \( f(x) \) gets closer to 5” means \( \lim_{x \to 3} f(x) = 5 \).
– A) \( f(3) = 5 \): Not necessarily true; the limit does not require \( f(3) \) to equal 5.
– B) \( f(5) = 3 \): Not true; the limit at \( x = 3 \) does not specify \( f(5) \).
– C) \( \lim_{x \to 3} f(x) = 5 \): This is the given property, so it must be true.
– D) \( \lim_{x \to 5} f(x) = 3 \): Not true; the limit at \( x = 5 \) is not specified.
Answer: C
▶️ Answer/Explanation
Solution
We need \( |f(x) – 7| < \varepsilon \) whenever \( |x – 2| < \delta \).
Given \( f(x) = 3x + 1 \), compute \( f(x) – 7 \):
\( f(x) – 7 = (3x + 1) – 7 = 3x – 6 = 3(x – 2) \).
So, \( |f(x) – 7| = 3|x – 2| \).
The condition becomes \( 3|x – 2| < \varepsilon \), or \( |x – 2| < \frac{\varepsilon}{3} \).
If \( |x – 2| < \delta \), then \( |f(x) – 7| < 3\delta \), and we need \( 3\delta \leq \varepsilon \), so \( \delta \leq \frac{\varepsilon}{3} \).
Evaluate the options:
– A) \( \delta = \frac{\varepsilon}{4} \): \( 3 \cdot \frac{\varepsilon}{4} = \frac{3\varepsilon}{4} < \varepsilon \), and \( \frac{\varepsilon}{4} \leq \frac{\varepsilon}{3} \), so this works.
– B) \( \delta = \frac{2}{\varepsilon} \): \( 3 \cdot \frac{2}{\varepsilon} = \frac{6}{\varepsilon} \), not \( < \varepsilon \) for all \( \varepsilon > 0 \) (e.g., \( \varepsilon = 1 \)).
– C) \( \delta = \frac{\varepsilon}{\varepsilon + 1} \): \( 3 \cdot \frac{\varepsilon}{\varepsilon + 1} < \varepsilon \) only if \( \varepsilon > 2 \), not for all \( \varepsilon \).
– D) \( \delta = \frac{\varepsilon + 1}{\varepsilon} \): \( 3 \cdot \frac{\varepsilon + 1}{\varepsilon} < \varepsilon \) only for large \( \varepsilon \), not all \( \varepsilon \).
– E) \( \delta = 3\varepsilon \): \( 3 \cdot 3\varepsilon = 9\varepsilon > \varepsilon \), does not work.
Answer: A