Now that’s a CAPTCHA. Reminds me, for some reason, of John Conway not allowing himself to log in unless he could correctly identify the day of the week of a given random date.
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It is tooooooooooo easy: ln(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…
But nice way to avooid bots…
ARCTG 1/0 = INFINITO
SEN 0 = 0
0 X INFINITO = 0
SOLUTION = LN2 = 0.69314718
0 x INFINITO != 0, be carefull 😉
Daniel’s solution is the correct one
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.
I hate to point it out to everyone commenting, but I think the title of this post kind of gave the game away.
i think that is not about the solution, it’s about how to get the solution, so in my opinion, the explanations are necessaries.