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Chapter 2: Problem 20

Use the precise definition of a limit to prove the following limits. $$\lim _{x \rightarrow 3}(-2 x+8)=2$$

Short Answer

Expert verified
Q: Prove the limit using the precise definition of a limit: $$\lim _{x \rightarrow 3}(-2 x+8)=2$$ A: To prove the given limit, we showed that for any given 蔚 > 0, there exists a corresponding 未 > 0 such that if 0 < |x - 3| < 未, then |(-2x + 8) - 2| < 蔚. We simplified the expression, found a relation between 蔚 and 未 by setting 未 = 蔚/2, and proved that the relation holds for any 蔚 > 0. Therefore, by the precise definition of a limit, $$\lim _{x \rightarrow 3}(-2 x+8)=2$$.

Step by step solution

01

Rewrite the Limit Statement using the Definitions of Limit, 蔚, and 未

Identify that our limit statement is of the form: $$\lim _{x \rightarrow 3}(-2 x+8)=2$$ Rewrite the limit statement using the precise definition of a limit: $$\forall \varepsilon > 0, \exists \delta > 0, \text{ such that if } 0 < |x - 3| < \delta, \text{ then } |(-2x + 8) - 2| < \varepsilon$$
02

Express the Function as an Absolute Value

Simplify the expression \(|(-2x + 8) - 2|\): $$|(-2x + 8) - 2| = |-2x + 6| = |2(3-x)| = 2|x-3|$$
03

Find the Relation between 蔚 and 未

We want \(2|x-3| < \varepsilon\) when \(0 < |x - 3| < \delta\). We can relate 蔚 and 未 by setting: $$2|x-3| < \varepsilon \Rightarrow |x-3| < \frac{\varepsilon}{2}$$ Therefore, we can choose \(\delta = \frac{\varepsilon}{2}\).
04

Prove That the Relation Holds

We have shown that if we choose \(\delta = \frac{\varepsilon}{2}\), then for any \(0 < |x-3| < \delta\), the following holds: $$|(-2x + 8) - 2| = 2|x-3| < 2\left(\frac{\varepsilon}{2}\right) = \varepsilon$$ This means that for any given 蔚 > 0, there exists a corresponding 未 > 0 (in this case, 未 = 蔚/2) such that if 0 < |x - 3| < 未, then |(-2x + 8) - 2| < 蔚. Thus, we have proved using the precise definition of a limit that: $$\lim _{x \rightarrow 3}(-2 x+8)=2$$

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