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The metric definition of a limit $$\lim_{x\rightarrow x_0} f(x)=\mathit{L}$$ is $$\forall\epsilon \; \exists\delta \; \forall x \; :\; 0<|x-x_0|<\delta\,\Longrightarrow\, |f(x)-\mathit{L}|<\epsilon$$ and we are using it in class to verify whether a certain limit is correct or not; however, I don't really understand the process of choosing a $\delta$ to verify the limit.

The classic example that my lecturer did in class is $$\lim_{x\rightarrow 3} x^2=9$$ and here is how he approached it: $$|x^2-9|<\epsilon \\ |x-3||x+3|<\epsilon \;\;\;\;\;\; and \;\;\;\;\;\;\;\; |x-3|<\delta\Rightarrow|x|<\delta+3 \\ |x-3|\,(|x|+3)<\epsilon \\ \delta\,(\delta+6)<\epsilon$$ then he limited delta to $\delta\le1$ so that $\delta\,(\delta+6)\le7\delta<\epsilon$. Finally, he put $$\delta=min\:\{\frac{\epsilon}{7};1\}$$

I don't really understand why he coud do this: $|x-3|<\delta\Rightarrow|x|<\delta+3$, because to me it looks like this would only be valid in certain cases. More generally speaking, I don't understand how much I can actually play around with the $|x-x_0|<\delta$, how I can modify it to be more useful and how to arrive then to a conclusion. If you also have any tips on how to approach this type of problems they are all welcome.

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We can assume wlog that $x\in (2,4)$, that is $\delta <1$, therefore $5<|x+3|<7$ and then

$$|x^2-9|=|x+3||x-3|<7|x-3| $$

therefore assuming $7|x-3|<\epsilon$ that is $|x-3|<\delta=\frac\epsilon 7$ we have

$$|x^2-9|<\epsilon $$

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