https://file.aiquickdraw.com/mathbot/file/1726464201453kzhzvyt8.jpg

This proof demonstrates that:

\[
\lim_{x \to 1} \frac{x}{1+x} = \frac{1}{2}
\]

by using the formal definition of a limit. Let's break down each step:

### Step-by-Step Breakdown:

1. **Statement of the Problem:**
   \[
   \lim_{x \to 1} \frac{x}{1+x} = \frac{1}{2}
   \]
   We want to prove this using the epsilon-delta definition of a limit.

2. **Assume an arbitrary \(\epsilon > 0\):**
   To apply the epsilon-delta definition, we need to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that:
   \[
   |x - 1| < \delta \quad \text{implies} \quad \left|\frac{x}{1+x} - \frac{1}{2}\right| < \epsilon.
   \]

3. **Transform the absolute difference:**
   Begin by manipulating the expression \(\left| \frac{x}{1+x} - \frac{1}{2} \right|\):
   \[
   \left| \frac{x}{1+x} - \frac{1}{2} \right| = \left| \frac{2x - (1+x)}{2(1+x)} \right| = \left| \frac{x-1}{2(1+x)} \right|.
   \]
   This step simplifies the expression to \(\frac{|x-1|}{2(1+x)}\).

4. **Bound the expression:**
   To find a bound for this expression, assume that \(|x - 1| \leq \frac{1}{2}\). This gives us bounds for \(x\):
   \[
   -\frac{1}{2} < x - 1 < \frac{1}{2} \quad \Rightarrow \quad \frac{1}{2} < x < \frac{3}{2}.
   \]
   Adding 1 to both sides:
   \[
   \frac{3}{2} < 1 + x < \frac{5}{2}.
   \]
   Therefore, we know that:
   \[
   \frac{1}{1+x} < \frac{2}{3}.
   \]

5. **Apply this bound:**
   Using this bound, we know:
   \[
   \frac{|x-1|}{2(1+x)} < \frac{|x-1|}{3}.
   \]
   This allows us to control the size of the original expression.

6. **Choose \(\delta\):**
   Now, we choose \(\delta = \min\left(\frac{1}{2}, 3\epsilon\right)\). This ensures that when \(|x - 1| < \delta\), the inequality
   \[
   \left| \frac{x}{1+x} - \frac{1}{2} \right| < \epsilon
   \]
   holds.

7. **Conclusion:**
   Hence, by the epsilon-delta definition of a limit, we have shown that:
   \[
   \lim_{x \to 1} \frac{x}{1+x} = \frac{1}{2}.
   \]

### Summary of the Proof Structure:

- We reduced the expression \(\left|\frac{x}{1+x} - \frac{1}{2}\right|\) to \(\frac{|x-1|}{2(1+x)}\).
- We bounded \(1 + x\) using the assumption \(|x - 1| \leq \frac{1}{2}\).
- We selected \(\delta\) based on the given \(\epsilon\) to ensure that the expression remains smaller than \(\epsilon\).
- Finally, we applied the delta condition to prove the limit using the epsilon-delta definition.

---

Would you like further details on any specific step?

Here are 5 related questions:
1. What is the epsilon-delta definition of a limit?
2. How do we determine bounds for variables in epsilon-delta proofs?
3. Why is bounding the expression \(\frac{|x-1|}{2(1+x)}\) necessary?
4. What is the importance of choosing \(\delta = \min\left(\frac{1}{2}, 3\epsilon\right)\)?
5. Can we apply a similar method to other types of limit problems?

**Tip:** In epsilon-delta proofs, simplifying the expression and finding manageable bounds is key to choosing an appropriate \(\delta\).

Explain this proof of \(\lim_{x \to 1} \frac{x}{1+x} = \frac{1}{2}\).

This page explains a step-by-step epsilon-delta proof for \(\lim_{x \to 1} \frac{x}{1+x} = \frac{1}{2}\). The proof simplifies the expression \(\frac{|x-1|}{2(1+x)}\) and demonstrates how to choose an appropriate delta to satisfy the epsilon condition.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation