The Quantum Pontiff

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1. It is tooooooooooo easy: ln(2)

   Comment by Eliatron — 6/4/2007 #

2. sin 1/0 is something between [-1,1], and arctg 0=0. Any value between -1 and 1, 0 times, is 0. Square root of 0 is 0…

   Comment by Daniel — 6/4/2007 #

3. But nice way to avooid bots…

   Comment by Daniel — 6/4/2007 #

4. ARCTG 1/0 = INFINITO
   SEN 0 = 0
   0 X INFINITO = 0
   SOLUTION = LN2 = 0.69314718

   Comment by GEORGE — 6/4/2007 #

5. 0 x INFINITO != 0, be carefull

   Daniel’s solution is the correct one

   Comment by Guiller — 6/4/2007 #

6. the solution is ln2, because, if the limit exists, it will be the sum of ln2 (a constant) and the lim(sqrt(arctg x . sin(1/x))), and this one is the sqrt(lim(arctg x. sin(1/x))), and this limit is 0, because, although lim(sin(1/x)) does not exist when x->0, is between -1 and 1, so this function, multiplied by another whith limit = 0 when x->0, as arctg x, has limit 0. So, finally, ln2+0=ln2.

   Comment by tanausú — 6/4/2007 #

7. I hate to point it out to everyone commenting, but I think the title of this post kind of gave the game away.

   Comment by Joe Fitzsimons — 6/4/2007 #

8. i think that is not about the solution, it’s about how to get the solution, so in my opinion, the explanations are necessaries.

   Comment by tanausú — 6/5/2007 #

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Dave Bacon
dabacon at cs dot washington dot edu

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