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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?


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up vote 3 down vote accepted

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 $|a|$.

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Umm I don't really understand the second part of your answer... Are you saying that there is a way we can set delta to be a form of epsilon? Wouldn't this be epsilon/3? – MathMathCookie Oct 19 '11 at 13:44
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
oops.. meant ε/(3a). And then let δ = min{ε/(3a),|a|}? – MathMathCookie Oct 19 '11 at 20:46
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

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