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I'm trying to study limits for my calculus class, but my textbook doesn't seem to be making much sense. In one of the examples, it shows us how to prove that $\lim_{x\to3} x^2 = 9$, with the following:

$|x^2-9| < \epsilon\text{ if }0<|x-3|<\delta$ $|x+3||x-3| < \epsilon \text{ if }0<|x-3|<\delta$ Assume $\delta\le1$, which then gives $-1.

I sort of get the above step, because that gives us a "boundary range" on either side of the limit that we're approaching. (Is that right?) However, the next step completely confuses me. The book draws the conclusion $5, and I have no idea how it gets that from the above. Can someone explain?

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    $|x-3|\lt 1$ means that $x$ is within $1$ of $3$. So it is between $2$ and $4$. Hence, $x+3$ is between $5$ and $7$ (since $x$ is somewhere between $2$ and $3$).2011-08-28

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Since you assumed $|x-3| < 1$, you have from this that

$ \Rightarrow -1 < x-3 <1 $ $\Rightarrow -1+6 < x-3+6 < 1+6$ $\Rightarrow 5 < x+3< 7$

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    And now I feel like an idiot; I guess I was looking too hard for a calculus-level answer. Thanks.2011-08-29