The Definition of the Limit — Question 6

Question 6

Let $$f(x) = \frac{1}{2}x + 4$$ Prove using the $\varepsilon$-$\delta$ definition of a limit that: $$\lim_{x \to -2} \left( \frac{1}{2}x + 4 \right) = 3$$

Question 6 - Solution

We are given: $$f(x) = \frac{1}{2}x + 4$$ and want to prove: $$\lim_{x \to -2} \left( \frac{1}{2}x + 4 \right) = 3$$ using the formal $\varepsilon$-$\delta$ definition of a limit.

Definition: For every $\varepsilon > 0$, there exists a $\delta > 0$ such that $$0 < |x + 2| < \delta \quad \Rightarrow \quad \left| \left( \frac{1}{2}x + 4 \right) - 3 \right| < \varepsilon$$

Step 1: Simplify the expression

$$\left| \left( \frac{1}{2}x + 4 \right) - 3 \right| = \left| \frac{1}{2}x + 1 \right| = \left| \frac{1}{2}(x + 2) \right| = \frac{1}{2}|x + 2|$$

So we want: $$\frac{1}{2}|x + 2| < \varepsilon \quad \Rightarrow \quad |x + 2| < 2\varepsilon$$

Step 2: Choose $\delta$

Let: $$\delta = 2\varepsilon$$

Step 3: Conclusion

Then whenever $0 < |x + 2| < \delta$, we have: $$\left| \left( \frac{1}{2}x + 4 \right) - 3 \right| = \frac{1}{2}|x + 2| < \frac{1}{2} \cdot \delta = \frac{1}{2}(2\varepsilon) = \varepsilon$$

Therefore, by the $\varepsilon$-$\delta$ definition, $$\boxed{\lim_{x \to -2} \left( \frac{1}{2}x + 4 \right) = 3}$$