Let's solve the problem step by step.

Part 1: Find the Limit

The function given is $f(x) = -5x + 3$. We need to find: $\lim_{x \to 2} f(x)$ Since $f(x)$ is a linear function, it's continuous everywhere, so we can find the limit by direct substitution: $\lim_{x \to 2} f(x) = f(2) = -5(2) + 3 = -10 + 3 = -7$ Thus, the real number $L$ is $-7$.

Part 2: Prove the Equality Using an $\varepsilon$-$\delta$ Proof

To prove $\lim_{x \to 2} f(x) = -7$ using an $\varepsilon$-$\delta$ proof, we proceed as follows:

1. Definition of the Limit: We need to show that for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|f(x) - (-7)| < \varepsilon$.

2. Set Up the Inequality: $|f(x) - (-7)| = |(-5x + 3) + 7| = |-5x + 10|$

   We can factor out the $-5$: $|-5x + 10| = 5|x - 2|$

3. Choosing $\delta$: We want $5|x - 2| < \varepsilon$. To make this true, set: $|x - 2| < \frac{\varepsilon}{5}$

   Therefore, let $\delta = \frac{\varepsilon}{5}$. This choice ensures that if $0 < |x - 2| < \delta$, then: $|f(x) - (-7)| = 5|x - 2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon$

4. Conclusion: By the definition of the limit, if $0 < |x - 2| < \delta$, then $|f(x) - (-7)| < \varepsilon$, confirming that: $\lim_{x \to 2} f(x) = -7$

This completes the $\varepsilon$-$\delta$ proof.

Would you like further details or explanations on any step?

Relative Questions

1. How do you find the limit of a polynomial function?
2. What makes a function continuous, and how does it relate to limits?
3. How do you approach proving a limit for more complex functions using the $\varepsilon$-$\delta$ method?
4. Can you explain how to choose $\delta$ when dealing with non-linear functions?
5. Why is it important that $\delta$ depends on $\varepsilon$ in an $\varepsilon$-$\delta$ proof?

Tip

When proving limits using $\varepsilon$-$\delta$, focus on manipulating the absolute value expression to isolate $|x - c|$. This step is crucial in finding a suitable $\delta$.