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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?

Thanks!

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1 Answer 1

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
1  
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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