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

We are tasked with proving that:

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

using the \(\epsilon\)-\(\delta\) definition of limits.

### The \(\epsilon\)-\(\delta\) Definition of a Limit

To prove that \(\lim_{x \to 0} f(x) = L\), we need to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - 0| < \delta\), it follows that:

\[
|f(x) - L| < \epsilon
\]

Here, the function is \(f(x) = \frac{4x + 5}{x + 1}\) and the limit \(L = 5\). So, we need to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that:

\[
\left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon \quad \text{whenever} \quad |x| < \delta.
\]

### Step-by-Step Proof

1. **Simplify the expression:**

   We begin by simplifying \(\left| \frac{4x + 5}{x + 1} - 5 \right|\):

   \[
   \frac{4x + 5}{x + 1} - 5 = \frac{4x + 5}{x + 1} - \frac{5(x + 1)}{x + 1} = \frac{(4x + 5) - 5(x + 1)}{x + 1}
   \]

   Simplifying the numerator:

   \[
   (4x + 5) - 5(x + 1) = 4x + 5 - 5x - 5 = -x
   \]

   Thus, the expression becomes:

   \[
   \frac{4x + 5}{x + 1} - 5 = \frac{-x}{x + 1}
   \]

   Therefore, we need to show that:

   \[
   \left| \frac{-x}{x + 1} \right| < \epsilon
   \]

2. **Bound the expression:**

   Now, let's bound \(\left| \frac{-x}{x + 1} \right|\):

   \[
   \left| \frac{-x}{x + 1} \right| = \frac{|x|}{|x + 1|}
   \]

   We want to find \(\delta > 0\) such that \(\frac{|x|}{|x + 1|} < \epsilon\) whenever \(0 < |x| < \delta\).

3. **Estimate \(|x + 1|\):**

   When \(x\) is close to 0, \(|x + 1|\) is close to 1. In fact, for small values of \(x\), say when \(|x| < 1\), we know that:

   \[
   |x + 1| \geq 1 - |x|
   \]

   So, for \(|x| < 1\), we have:

   \[
   \frac{|x|}{|x + 1|} \leq \frac{|x|}{1 - |x|}
   \]

4. **Find \(\delta\):**

   To ensure that \(\frac{|x|}{1 - |x|} < \epsilon\), we can choose \(\delta\) small enough. Specifically, we want:

   \[
   \frac{\delta}{1 - \delta} < \epsilon
   \]

   Solving for \(\delta\), we get:

   \[
   \delta < \frac{\epsilon}{1 + \epsilon}
   \]

5. **Conclusion:**

   Therefore, given any \(\epsilon > 0\), we can choose \(\delta = \frac{\epsilon}{1 + \epsilon}\). This ensures that whenever \(0 < |x| < \delta\), we have:

   \[
   \left| \frac{4x + 5}{x + 1} - 5 \right| < \epsilon
   \]

This completes the proof that \(\lim_{x \to 0} \frac{4x + 5}{x + 1} = 5\).

### Do you want details or have any questions?

Here are some related questions:
1. Can you prove \(\lim_{x \to 0} \frac{x^2}{x + 1} = 0\) using \(\epsilon\)-\(\delta\)?
2. How can we handle limits at infinity using \(\epsilon\)-\(\delta\) definitions?
3. What happens when you try \(\epsilon\)-\(\delta\) for discontinuous functions?
4. How is the concept of continuity tied to limits via \(\epsilon\)-\(\delta\)?
5. How does the \(\epsilon\)-\(\delta\) definition differ for multivariable limits?

**Tip:** When simplifying expressions involving limits, always aim to isolate the problematic variable to bound it effectively.

This step-by-step guide demonstrates how to prove that lim(x→0) (4x + 5)/(x + 1) = 5 using the epsilon-delta definition of limits. The proof involves simplifying expressions, bounding terms, and determining delta based on epsilon. Suitable for college-level calculus students.

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