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# 175

Valentines Day Card - February 14, 2010
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LOL! nice card!

Yeah, it's just the limit, isn't it!

But the scansion is a bit off. Lewis Carroll did better:

Yet what are all such gaieties to me
Whose thoughts are full of indices and surds?

x2 + 7x + 53 = 11/3

By the way, COULD you prove that? I tend to answer such questions with

"clearly, lim{x->2} x + 3 = 5. Q.E.D."

'cause it's obvious, but teachers have tried to prove stranger things to me...

OK, for every e>0, take d=e, so clearly 0<d<=e. Then if 0<|x-2|<d, we have |(x+3) - 5| < e. Therefore, the limit{x->2} x+3 = 5.

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