微積分...不定形之三角函數求極限

Question

lim x->0+ (x+sinx)^tanx
^是次方
解答:1
我算到lim x->0+ e^{tanx ln(x+sinx)}
求lim x->0+ tanx*ln(x+sinx)
=>lim x->0+ ln(x+sinx)/cotx...為(無窮/無窮) 然後羅必達後就不會了...@@

Answer 1

lim x->0 e^{tanx ln(x+sinx)}
先求lim x->0 tanx*ln(x+sinx)
=>lim x->0 ln(x+sinx)/cotx (無窮/無窮=> L'Hospital's rule)
=>lim x->0 [1+cosx)/(x+sinx)]/[- (sinx)]^2 ( 微分 cotx = [- (sinx)]^2 )
=>lim x->0 -[(sinx)^2*(1+cosx)]/(x+sinx) (0/0 => L'Hospital's rule)
=>lim x->0 [(sinx)^3-2(1+cosx)(sinx)(cosx)]/(1+cosx) = 0
=>lim x->0 tanx*ln(x+sinx) = 0
=> lim x->0 e^{tanx ln(x+sinx)} =1

2005-05-10 00:37:15 補充：
我打錯了,
微分 cotx = - 1/[(sinx)^2]