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:

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

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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!