# How should one prove: $\lim_{x\to a}~ x^2 = a^2$? [Sketch of proof inside]

How should one prove: $\displaystyle\lim_{x\to a}~x^2 = a^2$?

In order to find a delta, I've split $|x^2-a^2| < ε$ into $|x+a||x-a| < ε$, but I noticed that I can't just divide both sides by $|x+a|$ and be done with because of the pesky $x$ term... So I'm guessing that we set $|x-a|$ be less than some value, and assume for a second that this could be delta in order to move on with the proof?

Can someone point me in the right direction, if so? What should this "some value" be?

Thanks!

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If you insist $|x - a| < \delta \leq |a|$, then $|x + a| = |x - a + 2a| \leq |x - a| + 2|a| \leq 3|a|$.
Can you come up with a further restriction on $\delta$ that ensures $|x + a||x - a| < \varepsilon$? Surely this further restriction will depend on both $\varepsilon$ and $|a|$.
Just $\varepsilon / 3$ isn't enough. You'll get $|x^2 - a^2| < |a|\varepsilon$, right? How do you fix this? Restrict $\delta$ further - that's really your only option. Do you see how? –  Hans Parshall Oct 19 '11 at 14:14
Yes, that should work. The order is really (1) fix $\varepsilon$ (2) choose $\delta$ (3) show $|x^2 - a^2| < \varepsilon$. –  Hans Parshall Oct 19 '11 at 21:07