Proving Epsilon-Delta Inequalities for Limits of Functions

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SUMMARY

The discussion focuses on proving the epsilon-delta inequality for limits of functions, specifically showing that if \(0<\delta<1\) and \(|x-3|<\delta\), then \(|x^{2}-9|<7\delta\). The transformation of the inequality \(|x^{2}-9|\) into the product \(|x+3||x-3|\) is a key step in the proof. By applying the triangle inequality and the conditions given, the relationship between \(|x-3|\) and \(|x+3|\) is established, leading to the conclusion that \(|x^{2}-9|<7\delta\) holds true.

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Homework Statement



Show that, if [tex]0<\delta<1[/tex] and [tex]|x-3|<\delta[/tex], then [tex]|x^{2}-9|<7\delta[/tex]



Homework Equations





The Attempt at a Solution



I know that I need to transform one of the inequalities into the form of the other to prove it, but I don't see how. I can plug in values and of course it works since the limit of x^2 as x approaches 3 is 9, but I don't see how to show it algebraically. Please help
 
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|x²-9| = |x+3||x-3|<|x+3|d

From there, how can you use the triangle inequality and the hypothesis of |x-3|<d and 0<d<1 to get the result?
 
Note that if |x-3|< [itex]\delta[/itex] and and [itex]\delta[/itex]< 1, then |x-3|< 1 so -1< x-3< 1. What does that tell you about x+ 3 and |x+3|?
 

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