十万火急 谁能帮我解下这两个求极限的题目啊

Question

lim ex2-cosx
------------
x→0 xsinx

lim x 1
(—— - —— )
x→1 x-1 lnx

Answer 1

1.求极限x→0lim[(e^x²-cosx)/xsinx]
解：原式=x→0lim[(2xe^x²+sinx)/(sinx+xcosx)]=x→0lim[(2e^x²+4x²e^x²+cosx)/(cosx+cosx-xsinx)]
=x→0lim[(2e^x²+4x²e^x²+cosx)/(2cosx-xsinx)]=3/2
2.求极限x→1lim[x/(x-1)-1/lnx]
解：原式=x→1lim[(xlnx-x+1)/(x-1)lnx]=x→1lim[(lnx)/[lnx+(x-1)/x]=x→1lim[(1/x)/(1/x+1/x²)]=1/2

Answer 2

第一题 x→0 lim [(e^x²-cosx)/xsinx]
=x→0 lim [1+x²-(1-x²/2)]/x²
=x→0 lim 3x²/2/x²
=1.5
或
=x→0 lim [(2xe^x²+sinx) / (sinx+xcosx)]
=x→0 lim [(2e^x²+4x²e^x²+cosx) / (cosx+cosx-xsinx)]
=x→0 lim [(2e^x²+4x²e^x²+cosx) / (2cosx-xsinx)]
=1.5
第二题 x→1lim[x/(x-1)-1/lnx]
=x→1 lim [(xlnx-x+1)/(x-1)lnx]
=x→1 lim [(lnx) / [lnx+(x-1)/x]
=x→1 lim [(1/x)/(1/x+1/x²)]
=0.5

Answer 3

lim[e^(x^2) -cosx]/(xsinx)
=lim (1+x^2-cosx) /x^2
=lim(2sin^2 (x/2) +x^2) /x^2
=lim 3x^2/x^2
=3

lim[x/(x-1)-1/lnx]
=lim (xlnx-x+1) / [(x-1)lnx]
=lim (lnx+1-1) / (lnx+1-1/x)
=lim(1/x)/(1/x+1/x^2)
=1/2