Comments on: Ln 2 http://dabacon.org/pontiff/?p=1535 Theoretical Musings Thu, 09 Sep 2010 14:06:56 +0000 hourly 1 http://wordpress.org/?v=3.0.1 By: tanausú http://dabacon.org/pontiff/?p=1535#comment-183377 tanausú Tue, 05 Jun 2007 21:44:35 +0000 http://dabacon.org/pontiff/?p=1535#comment-183377 i think that is not about the solution, it's about how to get the solution, so in my opinion, the explanations are necessaries. i think that is not about the solution, it’s about how to get the solution, so in my opinion, the explanations are necessaries.

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]]> By: Joe Fitzsimons http://dabacon.org/pontiff/?p=1535#comment-183151 Joe Fitzsimons Mon, 04 Jun 2007 21:00:43 +0000 http://dabacon.org/pontiff/?p=1535#comment-183151 I hate to point it out to everyone commenting, but I think the title of this post kind of gave the game away. I hate to point it out to everyone commenting, but I think the title of this post kind of gave the game away.

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]]> By: tanausú http://dabacon.org/pontiff/?p=1535#comment-183109 tanausú Mon, 04 Jun 2007 16:23:35 +0000 http://dabacon.org/pontiff/?p=1535#comment-183109 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. 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.

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]]> By: Guiller http://dabacon.org/pontiff/?p=1535#comment-183095 Guiller Mon, 04 Jun 2007 15:28:55 +0000 http://dabacon.org/pontiff/?p=1535#comment-183095 0 x INFINITO != 0, be carefull ;) Daniel's solution is the correct one 0 x INFINITO != 0, be carefull ;)

Daniel’s solution is the correct one

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]]> By: GEORGE http://dabacon.org/pontiff/?p=1535#comment-183074 GEORGE Mon, 04 Jun 2007 13:35:29 +0000 http://dabacon.org/pontiff/?p=1535#comment-183074 ARCTG 1/0 = INFINITO SEN 0 = 0 0 X INFINITO = 0 SOLUTION = LN2 = 0.69314718 ARCTG 1/0 = INFINITO
SEN 0 = 0
0 X INFINITO = 0
SOLUTION = LN2 = 0.69314718

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]]> By: Daniel http://dabacon.org/pontiff/?p=1535#comment-183025 Daniel Mon, 04 Jun 2007 11:09:52 +0000 http://dabacon.org/pontiff/?p=1535#comment-183025 But nice way to avooid bots... But nice way to avooid bots…

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]]> By: Daniel http://dabacon.org/pontiff/?p=1535#comment-183024 Daniel Mon, 04 Jun 2007 11:08:05 +0000 http://dabacon.org/pontiff/?p=1535#comment-183024 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... 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…

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]]> By: Eliatron http://dabacon.org/pontiff/?p=1535#comment-183010 Eliatron Mon, 04 Jun 2007 09:43:48 +0000 http://dabacon.org/pontiff/?p=1535#comment-183010 It is tooooooooooo easy: ln(2) It is tooooooooooo easy: ln(2)

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