The Definition of the Limit — Question 5

Question 5

Let $$f(x) = 5x - 3$$ Prove, using the $\varepsilon$-$\delta$ definition of a limit, that: $$\lim_{x \to 2} (5x - 3) = 7$$

Question 5 - Solution

We are given: $$f(x) = 5x - 3$$ We want to prove: $$\lim_{x \to 2} (5x - 3) = 7$$ using the formal $\varepsilon$-$\delta$ definition of the limit.

Definition: For every $\varepsilon > 0$, there exists $\delta > 0$ such that $$0 < |x - 2| < \delta \quad \Rightarrow \quad |(5x - 3) - 7| < \varepsilon$$

Step 1: Simplify the expression

$$|(5x - 3) - 7| = |5x - 10| = 5|x - 2|$$

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

Step 2: Choose $\delta$

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

Step 3: Conclusion

Then whenever $0 < |x - 2| < \delta$, we have: $$|(5x - 3) - 7| = 5|x - 2| < 5\delta = \varepsilon$$

Therefore, by the $\varepsilon$-$\delta$ definition, $$\boxed{\lim_{x \to 2} (5x - 3) = 7}$$