prove that limit of 4x+5/x+1=5 as x approaches to 0 via value of delta to epsilon

To prove that

\[
\lim_{x \to 0} \frac{4x + 5}{x + 1} = 5
\]

using the **delta-epsilon definition** of a limit, we proceed as follows:

### Delta-Epsilon Definition of a Limit:
For any \(\epsilon > 0\), there must exist a \(\delta > 0\) such that for all \(x\),
\[
0 < |x - 0| < \delta \implies \left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon.
\]

Our goal is to find an expression for \(\delta\) in terms of \(\epsilon\).

### Step-by-step Proof:

1. **Start with the inequality:**
   \[
   \left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon.
   \]

2. **Simplify the expression inside the absolute value:**
   \[
   \frac{4x + 5}{x + 1} - 5 = \frac{4x + 5 - 5(x + 1)}{x + 1}.
   \]
   Simplifying further:
   \[
   = \frac{4x + 5 - 5x - 5}{x + 1} = \frac{-x}{x + 1}.
   \]

3. **Now, deal with the absolute value:**
   \[
   \left| \frac{-x}{x + 1} \right| = \frac{|x|}{|x + 1|}.
   \]

4. **Set up the inequality:**
   We need to ensure that:
   \[
   \frac{|x|}{|x + 1|} < \epsilon.
   \]

5. **Bound the denominator:**
   For \(x\) close to 0, the value of \(x + 1\) is close to 1. We can assume that for some small \(\delta\), say \(\delta < 1\), we have \(x + 1\) between 0.5 and 1.5. Therefore, \(|x + 1|\) can be bounded below by 0.5:
   \[
   |x + 1| \geq \frac{1}{2}.
   \]
   
6. **Apply the bound to the inequality:**
   \[
   \frac{|x|}{|x + 1|} < \epsilon \implies \frac{|x|}{\frac{1}{2}} < \epsilon \implies |x| < \frac{\epsilon}{2}.
   \]

7. **Conclusion:**
   Thus, if we choose \(\delta = \frac{\epsilon}{2}\), then for \(0 < |x| < \delta\), we have:
   \[
   \left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon.
   \]

### Final Statement:
By the delta-epsilon definition, since for every \(\epsilon > 0\), we can find \(\delta = \frac{\epsilon}{2}\) such that if \(0 < |x| < \delta\), then \(\left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon\), we have shown that:

\[
\lim_{x \to 0} \frac{4x + 5}{x + 1} = 5.
\]

---

Would you like further details or clarifications on this proof?

Here are 5 questions related to this topic:

1. How does the delta-epsilon definition apply to limits at infinity?
2. Can this technique be used for non-linear limits?
3. How does the choice of \(\delta\) change if \(\lim_{x \to 0} \frac{4x + 5}{x + 1} = 7\)?
4. What happens if \(x\) approaches values other than 0, like 2 or \(-1\)?
5. How does the presence of discontinuities affect delta-epsilon proofs?

**Tip:** When solving limits, always simplify the function first to make the delta-epsilon manipulation more manageable!

Learn how to prove the limit of (4x + 5)/(x + 1) equals 5 as x approaches 0 using the delta-epsilon definition. This detailed solution covers the delta-epsilon method, rational function simplifications, and the step-by-step process to find the appropriate delta for a given epsilon.

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