# Proving a Limit

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
Since you assumed $|x-3| < 1$, you have from this that
$$\Rightarrow -1 < x-3 <1$$ $$\Rightarrow -1+6 < x-3+6 < 1+6$$ $$\Rightarrow 5 < x+3< 7$$