Math Problem Statement
Given the function f(x) = {2x if x > -3, x^2 - 1 if x < -3}, find the limit of f(x) as x approaches -3.
Solution
We are given a piecewise function:
\begin{cases} 2x & \text{if } x > -3 \\ x^2 - 1 & \text{if } x < -3 \end{cases}$$ We need to evaluate the limit: $$\lim_{{x \to -3}} f(x)$$ ### Step 1: Evaluate the Left-Hand Limit $$\lim_{{x \to -3^-}} f(x)$$ For $$x < -3$$, the function is defined as $$f(x) = x^2 - 1$$. So, we compute: $$\lim_{{x \to -3^-}} f(x) = \lim_{{x \to -3^-}} (x^2 - 1) = (-3)^2 - 1 = 9 - 1 = 8$$ ### Step 2: Evaluate the Right-Hand Limit $$\lim_{{x \to -3^+}} f(x)$$ For $$x > -3$$, the function is defined as $$f(x) = 2x$$. Thus, we compute: $$\lim_{{x \to -3^+}} f(x) = \lim_{{x \to -3^+}} 2x = 2(-3) = -6$$ ### Step 3: Conclusion Since the left-hand limit $$\lim_{{x \to -3^-}} f(x) = 8$$ and the right-hand limit $$\lim_{{x \to -3^+}} f(x) = -6$$, the two limits are not equal. Therefore, the limit does not exist at $$x = -3$$. The final result is: $$\lim_{{x \to -3}} f(x) \text{ does not exist.}$$ Would you like further details or explanations? Here are some related questions to expand on the topic: 1. What is the significance of one-sided limits? 2. How do piecewise functions behave at boundary points? 3. Can a limit exist if the left-hand and right-hand limits are different? 4. How can discontinuities in a function be classified? 5. What is the difference between a limit existing and a function being continuous? **Tip:** When calculating limits for piecewise functions, always check the left-hand and right-hand limits separately to ensure continuity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Limits
Piecewise Functions
Formulas
Left-hand limit (lim x -> -3^-) and Right-hand limit (lim x -> -3^+)
Theorems
Definition of a Limit
Existence of a Limit Theorem
Suitable Grade Level
Grades 11-12 or Calculus Level