1. **Problem:** Prove that $$\lim_{x \to 1} (-4x + 3) = -1$$ using the definition of limit ($\varepsilon-\delta$ definition). 2. **Definition:** The limit $$\lim_{x \to a} f(x) = L$$ means for every $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that if $$0 < |x - a| < \delta$$ then $$|f(x) - L| < \varepsilon$$. 3. **Apply to our function:** Here, $$f(x) = -4x + 3$$, $$a = 1$$, and $$L = -1$$. 4. **Calculate $$|f(x) - L|$$:** $$ |(-4x + 3) - (-1)| = |-4x + 3 + 1| = |-4x + 4| = 4|1 - x| $$ 5. **Set $$|f(x) - L| < \varepsilon$$:** $$ 4|1 - x| < \varepsilon \implies |1 - x| < \frac{\varepsilon}{4} $$ 6. **Choose $$\delta$$:** Let $$\delta = \frac{\varepsilon}{4}$$. 7. **Conclusion:** For every $$\varepsilon > 0$$, if $$0 < |x - 1| < \delta$$ then $$|f(x) - (-1)| < \varepsilon$$, proving the limit. **Final answer:** $$\lim_{x \to 1} (-4x + 3) = -1$$.