https://file.aiquickdraw.com/mathbot/file/1724582924591kgss83m8.jpg

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.

To solve the given problem, let's break it down.

We are given a piecewise function \( F(x) \) defined as:

\[
F(x) =
\begin{cases} 
\frac{x^3 + x^2 + 2x + 1}{x + 1}, & x \neq 1 \\
-4, & x = 1 
\end{cases}
\]

### 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.

Explore how to find the limits of piecewise functions at x = 1. This problem involves evaluating one-sided limits and determining the existence of the limit at a point where the function changes. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation