高数,求极限1.lim（x→0）（tanx-sinx）/x^32.lim(x→派/2)sinx^(tanx)3.函数y=

1.用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/x³
=lim[x→0] (sinx/cosx-sinx)/x³
=lim[x→0] (sinx-sinxcosx)/(x³cosx)
=lim[x→0] sinx(1-cosx)/(x³cosx)
=lim[x→0] sin³x(1-cosx)/(x³sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/(sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/[(1-cos²x)cosx]
=lim[x→0] (sinx/x)³·(1-cosx)/[(1+cosx)(1-cosx)cosx]
=lim[x→0] (sinx/x)³·1/[(1+cosx)cosx]
=1·1/(1+1)
=1/2
2..sn=n√n!/n
0

设f(n) = 1/(n^2 n 1) 2/(n^2 n 2) . n/(n^2 n n)), f(n)

1.用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/x³
=lim[x→0] (sinx/cosx-sinx)/x³
=lim[x→0] (sinx-sinxcosx)/(x³cosx)
=lim[x→0] sinx(1-cosx)/(x³cosx)
=lim[x→0] sin
...

第一题用泰勒展开
第二题用对数
第三题当x->0 时b=lim(x->0)f(x)
左右两边除以x并取∞极限得k = lim(x->∞) f(x) / x