8 Replies to “Ln 2”

  1. 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…

  2. 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.

  3. 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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