Math Problem Statement
Let f(x) = {x^2 + 1 if x < -1, sqrt(x + 1) if x >= -1}. Compute the following limits or state that they do not exist: 1) lim x->-1^+ f(x), 2) lim x->-1^- f(x).
Solution
The function is defined as follows:
\begin{cases} x^2 + 1 & \text{if } x < -1 \\ \sqrt{x + 1} & \text{if } x \geq -1 \end{cases}$$ ### 1. $$\lim_{x \to -1^+} f(x)$$ For this limit, we consider the function's behavior as $$x \to -1$$ from the right (i.e., for values of $$x \geq -1$$). The function for $$x \geq -1$$ is $$f(x) = \sqrt{x + 1}$$. $$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \sqrt{x + 1} = \sqrt{-1 + 1} = \sqrt{0} = 0$$ Thus, $$\lim_{x \to -1^+} f(x) = 0$$. ### 2. $$\lim_{x \to -1^-} f(x)$$ For this limit, we consider the function's behavior as $$x \to -1$$ from the left (i.e., for values of $$x < -1$$). The function for $$x < -1$$ is $$f(x) = x^2 + 1$$. $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (x^2 + 1) = (-1)^2 + 1 = 1 + 1 = 2$$ Thus, $$\lim_{x \to -1^-} f(x) = 2$$. ### Summary of Limits 1. $$\lim_{x \to -1^+} f(x) = 0$$ 2. $$\lim_{x \to -1^-} f(x) = 2$$ Since the left-hand and right-hand limits do not agree, the overall limit $$\lim_{x \to -1} f(x)$$ does not exist. Do you want further details on any step? Here are 5 related questions: 1. What is the definition of one-sided limits? 2. How does the behavior of piecewise functions affect limits? 3. Can you explain why the limit does not exist if the one-sided limits are unequal? 4. How would the limit change if the function were continuous at $$x = -1$$? 5. Can we use L'Hopital's Rule to solve limits involving square roots? **Tip:** When dealing with piecewise functions, always check the left-hand and right-hand limits separately before concluding about the overall limit.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
One-Sided Limits
Formulas
Limit of sqrt(x + 1)
Limit of x^2 + 1
Theorems
Limits and Continuity
Definition of One-Sided Limits
Suitable Grade Level
Grades 10-12
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