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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 $-1<x-3<1$.

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 $5<x+3<7$, and I have no idea how it gets that from the above. Can someone explain?

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add 6 to both sides of the inequality. – Damien Aug 28 '11 at 23:37
$|x-3|\lt 1$ 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 $3$). – Arturo Magidin Aug 28 '11 at 23:52
up vote 8 down vote accepted

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. – Anonymous Aug 29 '11 at 0:04

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