作者LuisSantos (^______^)
看板trans_math
標題Re: ﹝極限﹞麻煩大家看一下這一題...
時間Sun Mar 18 20:49:05 2007
※ 引述《weakaday (耗子神)》之銘言：
: 小弟初次po版，請見諒
: lim [1/x]^2-[cotx]^2=?
: x→0
: 有答案，是2/3
: 不過我不知道過程怎麼來的
: 幫忙解一下，3q~~
這一題是台大85年轉學考微積分(B) 填充題的第6題
1
lim (----- - (cotx)^2)
x→0 x^2
1 (cosx)^2
= lim (----- - ---------)
x→0 x^2 (sinx)^2
(sinx)^2 - (x^2)((cosx)^2)
= lim ----------------------------
x→0 (x^2)((sinx)^2)
(sinx + (x)(cosx))(sinx - (x)(cosx))
= lim --------------------------------------
x→0 (x^2)((sinx)^2)
sinx + (x)(cosx) sinx - (x)(cosx)
= lim (------------------)(------------------)
x→0 sinx (x^2)(sinx)
sinx + (x)(cosx) sinx - (x)(cosx)
= (lim ------------------)(lim ------------------)
x→0 sinx x→0 (x^2)(sinx)
cosx + cosx -(x)(sinx) cosx - cosx + (x)(sinx)
= (lim ----------------------)(lim -------------------------)
x→0 cosx x→0 (2x)(sinx) + (x^2)(cosx)
1+1-0 (x)(sinx)
= (-----)(lim ---------------------------)
1 x→0 (2x)(sinx) + (x^2)(cosx)
sinx + (x)(cosx)
= (2)(lim ---------------------------------------------------)
x→0 (2)(sinx) + (2x)(cosx) + (2x)(cosx) - (x^2)(sinx)
sinx + (x)(cosx)
= (2)(lim --------------------------------------)
x→0 (2)(sinx) + (4x)(cosx) - (x^2)(sinx)
cosx + cosx - (x)(sinx)
= (2)(lim ---------------------------------------------------------------)
x→0 (2)(cosx) + (4)(cosx) - (4x)(sinx) - (2x)(sinx) - (x^2)(cosx)
2cosx - (x)(sinx)
= (2)(lim --------------------------------------)
x→0 (6)(cosx) - (6x)(sinx) - (x^2)(cosx)
2 - 0
= (2)(-----------)
6 - 0 - 0
2 2
= (2)(---) = ---
6 3
--
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