L ' Hospital Rule

👨🏻‍🏫 學術台數學

Green1114

2009-08-09 7:39 pm

lim x-> 0 (xln|tanx|)

ans : 0

更新1:

(1/tanx)(secx)^2 = 1/cosxsinx , is it bounded ?

更新2:

Why ln0 / infinite can apply the L ' hospital 's rule

回答 (1)

myisland8132

2009-08-09 9:11 pm

✔ 最佳答案

lim x-> 0+ (xln|tanx|)
=lim x-> 0+ (ln (tanx))/(1/x)
=lim x-> 0+ (1/tanx)(secx)^2/(1/x^2)
=lim x-> 0+ (x^2sinx)/(cosx)^3
= 0
Similar for lim x-> 0- (xln|tanx|)
So lim x-> 0 (xln|tanx|) = 0

2009-08-09 13:12:31 補充：
=lim x-> 0+ (1/tanx)(secx)^2/(1/x^2) missing the minus sign but the answer is still 0

👍 0 👎 0

收錄日期: 2021-04-26 13:44:59
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