The Definition of the Limit — Question 9

Question 9 - Solution

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

We want to show that: $$\forall \varepsilon > 0, \exists \delta > 0 \text{ such that if } 0 < |x - 2| < \delta, \text{ then } |f(x) - 9| < \varepsilon$$

Step 1: Work with the expression $|f(x) - 9|$ $$|f(x) - 9| = |5x - 1 - 9| = |5x - 10| = 5|x - 2|$$

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

Step 2: Choose $\delta$

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

Step 3: Conclusion

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

Therefore, by the $\varepsilon$-$\delta$ definition of the limit: $$\boxed{\lim_{x \to 2} (5x - 1) = 9}$$