1. **Problem 1:** Prove that $$\lim_{x \to 1} \frac{6 + 4x}{5} = 2$$ using the $$\varepsilon, \delta$$ definition of a limit. 2. Given $$\varepsilon > 0$$, we want to find $$\delta > 0$$ such that if $$0 < |x - 1| < \delta$$, then $$\left| \frac{6 + 4x}{5} - 2 \right| < \varepsilon$$. 3. Simplify the expression inside the absolute value: $$\left| \frac{6 + 4x}{5} - 2 \right| = \left| \frac{6 + 4x - 10}{5} \right| = \left| \frac{4x - 4}{5} \right| = \frac{4}{5} |x - 1|$$ 4. We want $$\frac{4}{5} |x - 1| < \varepsilon$$, which implies $$|x - 1| < \frac{5}{4} \varepsilon$$ 5. Choose $$\delta = \frac{5}{4} \varepsilon$$. Then if $$0 < |x - 1| < \delta$$, it follows that $$\left| \frac{6 + 4x}{5} - 2 \right| = \frac{4}{5} |x - 1| < \frac{4}{5} \delta = \frac{4}{5} \cdot \frac{5}{4} \varepsilon = \varepsilon$$ 6. Therefore, by the $$\varepsilon, \delta$$ definition, $$\lim_{x \to 1} \frac{6 + 4x}{5} = 2$$. --- 1. **Problem 2:** Prove that $$\lim_{x \to 4} \frac{x^2 + x - 20}{x - 4} = 9$$ using the $$\varepsilon, \delta$$ definition. 2. Given $$\varepsilon > 0$$, find $$\delta > 0$$ such that if $$0 < |x - 4| < \delta$$, then $$\left| \frac{x^2 + x - 20}{x - 4} - 9 \right| < \varepsilon$$. 3. Factor numerator: $$x^2 + x - 20 = (x + 5)(x - 4)$$ 4. Simplify the expression: $$\left| \frac{(x + 5)(x - 4)}{x - 4} - 9 \right| = |x + 5 - 9| = |x - 4|$$ for $$x \neq 4$$. 5. So the inequality becomes: $$|x - 4| < \varepsilon$$ 6. Choose $$\delta = \varepsilon$$. Then if $$0 < |x - 4| < \delta$$, it follows that $$\left| \frac{x^2 + x - 20}{x - 4} - 9 \right| = |x - 4| < \varepsilon$$ 7. Hence, by definition, $$\lim_{x \to 4} \frac{x^2 + x - 20}{x - 4} = 9$$. --- 1. **Problem 3:** Prove that $$\lim_{x \to 0} x^2 = 0$$ using the $$\varepsilon, \delta$$ definition. 2. Given $$\varepsilon > 0$$, find $$\delta > 0$$ such that if $$0 < |x - 0| < \delta$$, then $$|x^2 - 0| < \varepsilon$$. 3. Note that $$|x^2| = |x|^2$$ 4. We want $$|x|^2 < \varepsilon \implies |x| < \sqrt{\varepsilon}$$ 5. Choose $$\delta = \sqrt{\varepsilon}$$. Then if $$0 < |x| < \delta$$, it follows that $$|x^2| = |x|^2 < \delta^2 = \varepsilon$$ 6. Therefore, by the $$\varepsilon, \delta$$ definition, $$\lim_{x \to 0} x^2 = 0$$.