【問題】遇到難題了，誰能幫解

[steven7851 10]

誰能幫幫我～只是很基礎的數學罷了～

[rm2slg 10]

這...這是笑話嗎...= ="

[mapleaf 11]

因為是俄文所以比較難?(其實考信心)

此內容已被編輯, February 22, 2010 ,由 mapleaf

[onthewind 10]

lim後面的In是什麼意思阿??

[九天驚虹 10]

lim後面的In是什麼意思阿??

natural log (自然對數)

[steven7851 10]

因為是俄文所以比較難?(其實考信心)

感謝大大提供圖解

本人在其他網站提問時，尚得到其他人的解答，在此貼上以供參考

另解1：

lim {x->0} ln(2+(atan(x) * sin(1/x)) ** 1/2)

lim {x->0} ln(2+(atan(x) * (-1)) ** 1/2) less than or equal to lim {x->0} ln(2+(atan(x) * sin(1/x)) ** 1/2) less than or equal to lim {x->0} ln(2+(atan(x) * (1)) ** 1/2)

because lim {x->0} atan(x) = 0,

ln(2) less than or equal to lim {x->0} ln(2+(atan(x) * sin(1/2)) ** 1/2) less than or equal to ln(2)

so, the answer is ln2.

另解2：

because Ln only exists and is _continuous_ over (0,infinity), if a solution exists then, by the definition of continuity (dont feel like proving ln is continuous over that domain)...

lim Ln(2+sqrt(arctan(x)+sin(1/x)) is congruant to Ln( lim (2+ sqrt(arctan(x)*sin1/x))

this properties of limits allow this to be split up as

Ln(lim 2 + lim sqrt(arctan(x)*sin1/x))

since the limit of 2 is 2 this can be simplified as

Ln(2 + lim sqrt(arctan(x)+sin1/x)

to simplify i am going to just look at the "lim sqrt(arctan(x)*sin(1/x))" part for a bit

because sqrt is continuous ( assuming were looking at only the positive part)

lim sqrt(arctan(x)*sin(1/x)) is congruant to sqrt( lim arctan(x)*sin1/x))

for all x!=0, sin(1/x) is in [-1,1] so therefore using continuity of arctan and a bit of handwaving (waves hands)

that limit is 0 and working backwords we do indeed get the solution, Ln(2). The fact that sin(1/x) is undefined at 0 and that lim x->0 sin(1/x) is undefined does not matter because for all nonzero x, sin(1/x) is in [-1,1] and thus arctan(x)*sin(1/x) is bounded by magnitude of arctan(x) which goes to zero and so the limit goes to zero (squeeze theorem).

Note:

arctan(x)*sin(1/x) will have a negative value in any arbitrarily small interval around 0 so in actuality lim x->0 of sqrt(arctan(x)*sin(1/x)) is undefined, in the most rigorous sense: http://en.wikipedia.org/wiki/Limit_of_a_function#Functions_on_the_real_line .

So the actual soution is that that limit is undefined in the real line, but I think it would be Ln(2) in the complex plane. If I get bored at work tomorrow maybe I'll attempt it...

[論壇帳號 10]

這...這是笑話嗎...= ="

笑...笑話?

[rm2slg 10]

笑...笑話?

（我的意思是...怎會有這麼麻煩的認證碼...:| ）