The Definition of the Limit — Question 7

Question 7

Let $$f(x) = 3x - 1$$ Use the $\varepsilon$-$\delta$ definition of a limit to prove that: $$\lim_{x \to 2} f(x) = 5$$

Question 7 - Solution

We are given the function: $$f(x) = 3x - 1$$ and we are asked to prove: $$\lim_{x \to 2} (3x - 1) = 5$$ using the $\varepsilon$-$\delta$ definition.

Step 1: Recall the definition.

We must show that for every $\varepsilon > 0$, there exists a $\delta > 0$ such that $$0 < |x - 2| < \delta \quad \Rightarrow \quad |(3x - 1) - 5| < \varepsilon$$

Step 2: Simplify the expression inside the absolute value. $$|(3x - 1) - 5| = |3x - 6| = 3|x - 2|$$

So we require: $$3|x - 2| < \varepsilon \quad \Rightarrow \quad |x - 2| < \frac{\varepsilon}{3}$$

Step 3: Choose $\delta$.

Let $$\delta = \frac{\varepsilon}{3}$$

Step 4: Verification.

If $0 < |x - 2| < \delta = \frac{\varepsilon}{3}$, then: $$|f(x) - 5| = |3x - 6| = 3|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$$

Conclusion: By the $\varepsilon$-$\delta$ definition, we have: $$\boxed{\lim_{x \to 2} (3x - 1) = 5}$$