1、lim ln(1+x+2x^2)+ln(1-x

1、lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx

1个回答

lim(x→0) (ln cosx)/[ln(1+x^2)] 等于多少?
lim(x->0)[cosx-e^(-x^2/2)]/[x^2[x+ln(1-x)]]
lim (cosx)^ln(1/1+x^2)怎么算
lim/x-0(cosx)/ln(1+x^2)求极限
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x→0时 lim[1/ln(1+x^2)-1/(sinx)^2] =lim[(sinx)^2-ln(1+x^2)]/(s
x→0时lim(1-cosx/sin2x×ln(1+3x))
lim x→0[ln(cosx)]/x^2
lim(x→0)[(secx)^2-1] /(1-cosx)
lim x→0(1+2cosx)^secx