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.