1. **State the problem:** Prove that $$\lim_{x \to 5} \left(1 + \frac{1}{5}x\right) = 2$$ using the $\varepsilon, \delta$ definition of a limit. 2. **Recall the definition:** For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 5| < \delta$, then $$\left|\left(1 + \frac{1}{5}x\right) - 2\right| < \varepsilon.$$ This means the function values get arbitrarily close to 2 as $x$ approaches 5. 3. **Start with the expression inside the absolute value:** $$\left|\left(1 + \frac{1}{5}x\right) - 2\right| = \left|\frac{1}{5}x - 1\right|.$$ Rewrite it as: $$\left|\frac{1}{5}x - 1\right| = \frac{1}{5}|x - 5|$$ 4. **Set the inequality for $\varepsilon$:** We want $$\frac{1}{5}|x - 5| < \varepsilon.$$ Multiply both sides by 5: $$|x - 5| < 5\varepsilon.$$ 5. **Choose $\delta$:** Let $$\delta = 5\varepsilon.$$ Then whenever $$0 < |x - 5| < \delta,$$ we have $$\left|\left(1 + \frac{1}{5}x\right) - 2\right| = \frac{1}{5}|x - 5| < \frac{1}{5} \delta = \frac{1}{5} \times 5\varepsilon = \varepsilon.$$ 6. **Conclusion:** By the $\varepsilon, \delta$ definition of a limit, since for every $\varepsilon > 0$ we can find $\delta = 5\varepsilon$ such that $0 < |x - 5| < \delta$ implies $$\left|\left(1 + \frac{1}{5}x\right) - 2\right| < \varepsilon,$$ we conclude $$\lim_{x \to 5} \left(1 + \frac{1}{5}x\right) = 2.$$