3. Case 1 - Real Limit

Case 1

Find the $\delta$ & $\epsilon$ that satisfies

$$\lim_{x \to 1} 5x + 2 = 7$$

Answer

Step 1: Express $|f(x) - L| < \epsilon$ in terms of $x$

$$|f(x) - L| \; = \; |(5x + 2) - 7| \; = \; |5(x - 1)|$$

Step 2: Relate it to $\epsilon$

$$|f(x) - L| < \epsilon \quad \to \quad |5(x - 1)| < \epsilon$$

Step 3: Solve for $\epsilon$

$$|x - 1| < \frac{\epsilon}{5}$$

Step 4: Choose $\delta$

$$\delta = \frac{\epsilon}{5}$$

Note: saying $|x - 1| < \dfrac{\epsilon}{n} \quad$ where $\quad n \in \mathbb{R}, \; n \geq 3 \quad$ so $\quad \delta > \dfrac{\epsilon}{n} \quad$ is also true