▶️ Answer/Explanation
Solution
Correct Answer: C
Solution:
For the right-hand limit (x→2⁺), we use the x > 2 case:
f(x) = x² – 4x + 8
Evaluating the limit:
\(\lim_{x \to 2^{+}} (x^2 – 4x + 8) = 2^2 – 4(2) + 8 = 4 – 8 + 8 = 4\)
Key Concept:
Right-hand limits only consider the function definition for values approaching from above the point.
▶️ Answer/Explanation
Solution
Correct Answer: C
Step-by-Step Solution:
1. Break down the limit using limit laws:
\(\lim_{x \to 2}(h(x)(5f(x)+g(x))) = \lim_{x \to 2}h(x) \times (5\lim_{x \to 2}f(x) + \lim_{x \to 2}g(x))\)
2. Substitute values from the table:
\(= 2 \times (5 \times 4 + (-6))\)
\(= 2 \times (20 – 6)\)
\(= 2 \times 14 = 28\)
Key Concept:
The limit of a product/sum can be evaluated as the product/sum of limits when all individual limits exist. Note that function values (f(2), g(2), h(2)) are irrelevant when the limits exist.
▶️ Answer/Explanation
Solution
Correct Answer: C
Factor numerator and denominator:
\(\frac{x^{2}-9}{x^2-2x-15} = \frac{(x+3)(x-3)}{(x+3)(x-5)}\)
\(= \frac{x-3}{x-5}\) for x ≠ -3
Evaluate the limit:
\(\lim_{x \to -3} \frac{x-3}{x-5} = \frac{-6}{-8} = \frac{3}{4}\)
▶️ Answer/Explanation
Solution
Correct Answer: C
Simplify the function:
\(f(x) = \frac{(x-2)(x+2)}{(x+3)(x-2)}\)
\(= \frac{x+2}{x+3}\) for x ≠ 2
Evaluate the limit:
\(\lim_{x \to 2} \frac{x+2}{x+3} = \frac{4}{5}\)
Key Concept:
Simplifying rational functions before evaluating limits
▶️ Answer/Explanation
Solution
Correct Answer: B
Factor the denominator:
\(\frac{x+4}{x^3-16x} = \frac{x+4}{x(x-4)(x+4)}\)
\(= \frac{1}{x(x-4)}\) for x ≠ -4
Evaluate the limit:
\(\lim_{x \to -4} \frac{1}{x(x-4)} = \frac{1}{-4(-4-4)} = \frac{1}{32}\)