Math Problem Statement
Given f(x) = {2x, x > 4; 5, x = 4; x^2 - 8, x < 4}, find lim(x→4) f(x) if it exists.
Solution
We are asked to find , given the following piecewise function:
\begin{cases} 2x, & x > 4 \\ 5, & x = 4 \\ x^2 - 8, & x < 4 \end{cases}$$ ### Step 1: Evaluate the Left-Hand Limit $$\lim_{{x \to 4^-}} f(x)$$ This is the limit of $$f(x)$$ as $$x$$ approaches 4 from the left ($$x < 4$$). For $$x < 4$$, the function is $$f(x) = x^2 - 8$$. So, we calculate: $$\lim_{{x \to 4^-}} f(x) = \lim_{{x \to 4^-}} (x^2 - 8) = 4^2 - 8 = 16 - 8 = 8$$ ### Step 2: Evaluate the Right-Hand Limit $$\lim_{{x \to 4^+}} f(x)$$ This is the limit of $$f(x)$$ as $$x$$ approaches 4 from the right ($$x > 4$$). For $$x > 4$$, the function is $$f(x) = 2x$$. So, we calculate: $$\lim_{{x \to 4^+}} f(x) = \lim_{{x \to 4^+}} 2x = 2(4) = 8$$ ### Step 3: Check if the Two-Sided Limit Exists Since both the left-hand limit and the right-hand limit are equal to 8, the two-sided limit exists. Therefore: $$\lim_{{x \to 4}} f(x) = 8$$ Thus, the limit exists and its value is 8. Would you like more details or have any questions? Here are five follow-up questions that you might find useful: 1. What if the left-hand and right-hand limits were not equal? What would the conclusion be? 2. How does the value of $$f(4)$$ affect the limit $$\lim_{{x \to 4}} f(x)$$? 3. What does it mean for a limit to "not exist" in a formal sense? 4. Can a piecewise function have a continuous limit at a boundary point like $$x = 4$$? 5. How do you compute limits for more complicated piecewise functions? **Tip:** When calculating limits for piecewise functions, always compute the left-hand and right-hand limits separately before drawing any conclusions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Left-hand Limit
Right-hand Limit
Formulas
Limit of a function
lim(x→c) f(x)
Left-hand and right-hand limits
Theorems
Two-Sided Limit Theorem
Continuity of piecewise functions
Suitable Grade Level
Grades 11-12 (High School Calculus or Pre-Calculus)