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Chapter 2: Problem 24

Use the precise definition of a limit to prove the following limits. $$\left.\lim _{x \rightarrow 3}(x-3)^{2}=0 \text { (Hint: Use the identity } \sqrt{x^{2}}=|x| .\right)$$

Short Answer

Expert verified
Question: Use the precise definition of a limit to prove that $$\lim _{x \rightarrow 3}(x-3)^{2}=0$$ Answer: Choose $\delta = \sqrt{\epsilon}$. When $0 < |x - 3| < \delta$, then $|(x-3)^{2} - 0| < \epsilon$.

Step by step solution

01

State the Precise Definition of a Limit

According to the precise definition of a limit, we need to show that for any given number \(\epsilon > 0\), there exists a number \(\delta > 0\) such that if \(0 < |x - 3| < \delta\), then \(|(x-3)^{2} - 0| < \epsilon\).
02

Begin the proof with the expression\(|(x-3)^{2} - 0|\)

We are given the hint \(\sqrt{x^{2}}=|x|\). Apply this to our expression: \(|(x-3)^{2} - 0| = |x-3| \cdot |x-3|\).
03

Choose the correct \(\delta\)

Set \(\delta\) to be equal to \(\sqrt{\epsilon}\). We can write this as \(\delta = \sqrt{\epsilon}\). So now we have: \(0 < |x - 3| < \delta = \sqrt{\epsilon}\).
04

Substitute \(\delta\) and prove the limit

We can now substitute the value of \(\delta\) in our expression from Step 2: \(|(x-3)^{2} - 0| = |x-3| \cdot |x-3| < \sqrt{\epsilon} \cdot \sqrt{\epsilon}\). Solving this inequality, we get: \(|(x-3)^{2} - 0| < \epsilon\).
05

Conclusion

Since we were able to show that for any given \(\epsilon > 0\), there exists a number \(\delta = \sqrt{\epsilon}\) such that if \(0 < |x - 3| < \delta\), then \(|(x-3)^{2} - 0| < \epsilon\), we can conclude that the limit exists and $$\lim _{x \rightarrow 3}(x-3)^{2}=0$$. The proof is complete.

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Most popular questions from this chapter

Sketch a possible graph of a function \(f\) that satisfies all of the given conditions. Be sure to identify all vertical and horizontal asymptotes. $$f(-1)=-2, f(1)=2, f(0)=0, \lim _{x \rightarrow \infty} f(x)=1$$, $$\lim _{x \rightarrow-\infty} f(x)=-1$$

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