Re: 問一題極限
看板trans_math作者iamhido (.....................)時間21年前 (2004/10/25 15:05)推噓6(6推 0噓 1→)留言7則, 2人參與討論串2/2 (看更多)
※ 引述《pianoman (隨時都要快快樂樂的)》之銘言:
: 拜託高手一下
: limIn(sinx/x)
: x-o --------
: x 謝謝 拜託一下了 台大的眾高手
原式為(0/0),屬於標準不定型,可利用L'hospital
先將原式改寫為 lim ln(sinx)-lnx
x->0 ------------
x
= lim (cosx/sinx)-1/x
x->0 ---------------
1
= lim xcosx-sinx 0
x->0 ---------- ( - )
xsinx 0
= lim x(cosx-1)+(x-sinx)
x->0 ------------------
xsinx
= lim (cosx-1) + lim (x-sinx) x
x->0 -------- ------------‧----- ‧x
sinx x^3 sinx
= lim (-sinx) + (1/6)‧1‧0
x->0 -------
cosx
= 0
#
若有錯誤只指教。
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