Math Problem Statement
Let f(x) = { x^2 + 2 if x ≤ -1; 2x - 5 if x > -1 }. Find lim f(x) as x approaches -1.
Solution
We are asked to find:
where the function is defined piecewise as:
\begin{cases} x^2 + 2 & \text{for } x \leq -1, \\ 2x - 5 & \text{for } x > -1. \end{cases}$$ ### Steps to solve: #### 1. **Left-hand limit ($$x \to -1^-$$)**: Since $$x \to -1^-$$ approaches from the left (values of $$x \leq -1$$), we use the function $$f(x) = x^2 + 2$$. $$\lim_{x \to -1^-} f(x) = (-1)^2 + 2 = 1 + 2 = 3.$$ #### 2. **Right-hand limit ($$x \to -1^+$$)**: For $$x \to -1^+$$, approaching from the right (values of $$x > -1$$), we use $$f(x) = 2x - 5$$. $$\lim_{x \to -1^+} f(x) = 2(-1) - 5 = -2 - 5 = -7.$$ #### 3. **Conclusion**: The left-hand limit is 3, and the right-hand limit is -7. Since the left and right limits are not equal, the two-sided limit does not exist. Thus, the correct answer is: $$\boxed{d. \text{Does not exist}}$$ Would you like any further explanation or details? Here are 5 follow-up questions to deepen understanding: 1. What is the significance of left-hand and right-hand limits in piecewise functions? 2. Can a function be continuous if its two-sided limit does not exist? 3. How do we determine continuity at a point in a piecewise function? 4. What happens if the left-hand and right-hand limits are equal? 5. Could this piecewise function be modified to make the limit exist? **Tip:** Always check both the left-hand and right-hand limits when working with piecewise functions!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
Left-hand limit calculation: lim_{x→-1^-} f(x)
Right-hand limit calculation: lim_{x→-1^+} f(x)
Theorems
Existence of Limit Theorem
Suitable Grade Level
Grades 11-12 or Early College