Math Problem Statement
find (a) lim x→1+ F(x) and (b) lim x→1- F(x). In each case, determine if lim x→1 F(x) exists.
Solution
To solve the given problem, let's break it down.
We are given a piecewise function defined as:
\begin{cases} \frac{x^3 + x^2 + 2x + 1}{x + 1}, & x \neq 1 \\ -4, & x = 1 \end{cases}$$ ### Part (a): Finding $$\lim_{x \to 1^+} F(x)$$ For $$x \to 1^+$$ (the right-hand limit), we will use the first part of the piecewise function because $$x \neq 1$$. We need to compute: $$\lim_{x \to 1^+} \frac{x^3 + x^2 + 2x + 1}{x + 1}$$ To simplify, first factor the numerator $$x^3 + x^2 + 2x + 1$$: $$x^3 + x^2 + 2x + 1 = (x + 1)(x^2 + 1)$$ So the expression becomes: $$\frac{(x + 1)(x^2 + 1)}{x + 1}$$ For $$x \neq -1$$, this simplifies to: $$x^2 + 1$$ Now, evaluating the limit as $$x \to 1$$: $$\lim_{x \to 1^+} (x^2 + 1) = 1^2 + 1 = 2$$ Thus, $$\lim_{x \to 1^+} F(x) = 2$$. ### Part (b): Finding $$\lim_{x \to 1^-} F(x)$$ Similarly, for $$x \to 1^-$$, we still use the first part of the function: $$\lim_{x \to 1^-} \frac{x^3 + x^2 + 2x + 1}{x + 1}$$ As before, this simplifies to $$x^2 + 1$$, so: $$\lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2$$ Thus, $$\lim_{x \to 1^-} F(x) = 2$$. ### Does $$\lim_{x \to 1} F(x)$$ exist? Since both one-sided limits are equal: $$\lim_{x \to 1^+} F(x) = \lim_{x \to 1^-} F(x) = 2$$ This implies that $$\lim_{x \to 1} F(x) = 2$$ exists. ### Final Answer: - (a) $$\lim_{x \to 1^+} F(x) = 2$$ - (b) $$\lim_{x \to 1^-} F(x) = 2$$ - The limit $$\lim_{x \to 1} F(x)$$ exists and equals 2. Would you like a more detailed explanation of any part of this solution? --- Here are 5 related questions: 1. What is the definition of a piecewise function? 2. How do you find limits for piecewise functions at the points where the function changes? 3. What does it mean if the left-hand and right-hand limits are different? 4. How do you use factoring to simplify limit expressions? 5. What are common techniques for solving limits algebraically? **Tip:** Always check if a function is continuous at the point of interest when computing limits, as this can simplify your work.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Formulas
-
Theorems
-
Suitable Grade Level
Grades 11-12