The Definition of the Limit — Question 2

Question 2 - Solution

We are given: $$f(x) = x^2 + 3x, \quad \lim_{x \to 1} f(x) = 4$$

We want to prove using the $\varepsilon$-$\delta$ definition:

Goal: For every $\varepsilon > 0$, we must find $\delta > 0$ such that: $$0 < |x - 1| < \delta \quad \Rightarrow \quad |f(x) - 4| < \varepsilon$$

Step 1: Write the expression $|f(x) - 4|$: $$|f(x) - 4| = |x^2 + 3x - 4|$$

Factor the expression: $$x^2 + 3x - 4 = (x - 1)(x + 4) \Rightarrow |f(x) - 4| = |x - 1||x + 4|$$

We want: $$|x - 1||x + 4| < \varepsilon$$

Step 2: Estimate $|x + 4|$:

Assume $|x - 1| < 1 \Rightarrow x \in (0, 2) \Rightarrow x + 4 \in (4, 6) \Rightarrow |x + 4| < 6$

Then: $$|f(x) - 4| = |x - 1||x + 4| < 6|x - 1|$$

To make $|f(x) - 4| < \varepsilon$, it is enough to make: $$6|x - 1| < \varepsilon \Rightarrow |x - 1| < \frac{\varepsilon}{6}$$

Step 3: Let $\delta = \min\left\{1, \frac{\varepsilon}{6} \right\}$

Then if $0 < |x - 1| < \delta$, we have: $$|f(x) - 4| < \varepsilon$$

Conclusion: $$\text{Therefore, } \lim_{x \to 1} (x^2 + 3x) = 4 \text{ by the }\varepsilon\text{-}\delta\text{ definition.}$$