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# L ' Hospital Rule

lim x-> 0 (xln|tanx|)

ans : 0

Update:

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

Update 2:

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

### 1 Answer

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- myisland8132Lv 71 decade agoFavorite Answer
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

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