Green1114 asked in 科學及數學數學 · 1 decade ago

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

Rating
  • 1 decade ago
    Favorite 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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