The Definition of the Limit — Question 3

Question 3

Let $$f(x) = 5x - 1$$ Prove, using the $\varepsilon$-$\delta$ definition of a limit, that: $$\lim_{x \to 2} f(x) = 9$$

Question 3 - Solution

We are given: $$f(x) = 5x - 1, \quad \lim_{x \to 2} f(x) = 9$$

We will use the $\varepsilon$-$\delta$ definition of a limit:

Goal: For every $\varepsilon > 0$, find $\delta > 0$ such that: $$0 < |x - 2| < \delta \quad \Rightarrow \quad |f(x) - 9| < \varepsilon$$

Step 1: Express $|f(x) - 9|$: $$|f(x) - 9| = |5x - 1 - 9| = |5x - 10| = 5|x - 2|$$

We want: $$5|x - 2| < \varepsilon \quad \Rightarrow \quad |x - 2| < \frac{\varepsilon}{5}$$

Step 2: Choose $\delta$:

Let: $$\delta = \frac{\varepsilon}{5}$$

Then if $0 < |x - 2| < \delta$, we have: $$|f(x) - 9| = 5|x - 2| < 5\delta = \varepsilon$$

Conclusion: $$\text{Therefore, } \lim_{x \to 2} (5x - 1) = 9 \text{ by the }\varepsilon\text{-}\delta\text{ definition.}$$