Math Problem Statement
For the function f(x) shown below, determine lim(x → -2) f(x).
f(x) = { 10 + 5x for x < -2
x^2 - 4 for x ≥ -2 }
Solution
To determine , we need to analyze the piecewise function . The function is defined differently depending on whether is less than or greater than/equal to :
\begin{cases} 10 + 5x & \text{for } x < -2 \\ x^2 - 4 & \text{for } x \geq -2 \end{cases}$$ ### Step 1: Approach $$x = -2$$ from the left (left-hand limit) When $$x \to -2^-$$, we use the definition of $$f(x)$$ for $$x < -2$$: $$f(x) = 10 + 5x$$ Substituting $$x = -2$$: $$\lim_{x \to -2^-} f(x) = 10 + 5(-2) = 10 - 10 = 0$$ ### Step 2: Approach $$x = -2$$ from the right (right-hand limit) When $$x \to -2^+$$, we use the definition of $$f(x)$$ for $$x \geq -2$$: $$f(x) = x^2 - 4$$ Substituting $$x = -2$$: $$\lim_{x \to -2^+} f(x) = (-2)^2 - 4 = 4 - 4 = 0$$ ### Step 3: Compare the left-hand and right-hand limits Since both the left-hand limit and right-hand limit as $$x \to -2$$ are equal to 0, we conclude that the two-sided limit exists and is: $$\lim_{x \to -2} f(x) = 0$$ ### Conclusion: The limit of $$f(x)$$ as $$x$$ approaches $$-2$$ is $$0$$. --- Would you like more details or have any questions about this solution? Here are five related questions for further practice: 1. What is the limit of $$f(x)$$ as $$x \to -2$$ if $$f(x)$$ was defined differently for $$x \geq -2$$? 2. How do you find the limit of a piecewise function at a point where the definition changes? 3. What is the value of $$f(-2)$$ in this case? Is it continuous at $$x = -2$$? 4. What is the limit of $$f(x)$$ as $$x \to \infty$$? 5. How does the concept of one-sided limits help determine the existence of a two-sided limit? **Tip:** Always remember to compare the left-hand and right-hand limits to determine whether a two-sided limit exists.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Formulas
Limit definition
Substitution in limits
Theorems
Limit existence criteria
Piecewise function limit evaluation
Suitable Grade Level
Grades 11-12