Math Problem Statement
Consider the following function: f(x) = { 5/x, if x < 0; x^2 - 3x, if 0 ≤ x < 4; 3x, if x ≥ 4 }. Find: a. lim(x→0-) f(x); b. lim(x→0+) f(x); c. lim(x→3) f(x).
Solution
Let's analyze the function and compute the requested limits step-by-step. The piecewise function is:
\frac{5}{x}, & \text{if } x < 0 \\ x^2 - 3x, & \text{if } 0 \leq x < 4 \\ 3x, & \text{if } x \geq 4 \end{cases}$$ ### Part (a): $$\lim_{x \to 0^-} f(x)$$ For $$x \to 0^-$$, we are approaching zero from the left side (where $$x < 0$$). In this case, the relevant piece of the function is $$f(x) = \frac{5}{x}$$. $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{5}{x} = -\infty$$ Because as $$x$$ approaches 0 from the negative side, $$\frac{5}{x}$$ tends to $$-\infty$$. ### Part (b): $$\lim_{x \to 0^+} f(x)$$ For $$x \to 0^+$$, we are approaching zero from the right side (where $$x \geq 0$$). In this case, the relevant piece of the function is $$f(x) = x^2 - 3x$$. $$\lim_{x \to 0^+} f(x) = (0)^2 - 3(0) = 0$$ ### Part (c): $$\lim_{x \to 3} f(x)$$ At $$x = 3$$, we use the middle piece of the function, $$f(x) = x^2 - 3x$$, because $$0 \leq x < 4$$. $$f(3) = (3)^2 - 3(3) = 9 - 9 = 0$$ Thus, the limit is: $$\lim_{x \to 3} f(x) = 0$$ ### Final Answers: - (a) $$\lim_{x \to 0^-} f(x) = -\infty$$ - (b) $$\lim_{x \to 0^+} f(x) = 0$$ - (c) $$\lim_{x \to 3} f(x) = 0$$ Let me know if you'd like further details or explanations on any part of this. Here are five related questions to expand on: 1. What happens if $$f(x)$$ is evaluated at a specific point, such as $$x = -2$$? 2. How can we determine if a piecewise function is continuous at a point like $$x = 0$$? 3. What does it mean for a limit to not exist, and can this apply to any piecewise function? 4. Can a function have different limits from the left and right atAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Formulas
Limit definition for piecewise functions
Theorems
Definition of limit
Properties of limits
Suitable Grade Level
Grades 11-12