lim (1/x^2)∫[(1+2t)e^(t-x^2)]dt=?注：积分范围是0-》x^2 lim的下面是x-》00

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lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注：积分范围是0-》x^2 lim的下面是x-》00
lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注：积分范围是0-》x^2 lim的下面是x-》00

lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注：积分范围是0-》x^2 lim的下面是x-》00
=lim (1/x^2)*∫[(1+2(t-x^2 + x^2))*e^(t-x^2)]d(t-x^2)
令u=t-x^2,则u的范围是 -x^2 至 0.
原式=lim (1/x^2)*∫[(1+2(u + x^2))*e^u ]du
=lim (1/x^2)*{∫[(1 + 2x^2 + 2u )*e^u ]du }
=lim (1/x^2)*{∫(1 + 2x^2)e^u du + 2∫u *e^u du }
=lim {(1 + 2x^2)∫e^u du + 2∫u *e^u du } /x^2
=lim {(1 + 2x^2)e^u|< u从-x^2 至 0> + 2∫u *e^u du } /x^2
=lim {(1 + 2x^2)(1-e^(-x²) + 2∫u *e^u du ) } /x^2
=lim [(1 + 2x^2)/x^2]·lim (1-e^(-x²) + 2∫u *e^u du )
= 2·lim (1-0 + 2∫u *e^u du )
= 2·lim (1 + 2∫u de^u )
= 2·lim (1 + 2(u·e^u -∫e^u du ) )
= 2·lim (1 + 2(u-1)·e^u| )
= 2·lim (1 + 2[(0-1)·e^0) - (-x^2 -1)·e^(-x^2) ] )
= 2·lim (1 + 2[-1 + (x^2 +1)·e^(-x^2) ] )
= 2·lim (1 + 2[-1 + (x^2 +1)·e^(-x^2) ] )
= -2 + 4·lim (x^2 +1) / e^(x^2)
= -2 + 4·lim (2x) / [2x·e^(x^2)] 【洛比达法则】
= -2 + 4·lim 1 / e^(x^2)
= -2 + 0
= -2

定积分：(1) lim(x→a) 1/(x-a) ∫[a,x] f(t)dt(2) lim(x→∞) ∫[x,x+1] (sint)/t dt lim x→0[∫上x下0 cos(t^2)dt]/x ; lim x→0[∫上x下0 ln(1+t)dt]/(xsinx) lim (1/x^2)*∫[(1+2t)*e^(t-x^2)]dt=?注：积分范围是0-》x^2 lim的下面是x-》00 lim(x→o) (∫(0,x^2) √(1+t^2)dt) /x^2 lim x-0 (∫0-x^2 （ln（1+2t）dt)/x^4 lim(n,0)x/(1-e^x^2)∫(0,x)e^t^2dt lim(x→0)[x∫(x→1)(cost/t^2)dt] lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3 lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3 lim(x→0)(x∫(x→1)cost/t^2dt) lim(x→0)1/x∫[x,0](1+t^2)*e^(t^2-x^2)dt f(t)=lim x→无穷大 [t(1+1/x)^2tx] 求f'(t) lim x-0 ∫0-π （ln（1+2t）dt）/x^4 lim x-0 ∫0-π （ln（1+2t）dt）/x^4 lim（1+2t/x）^3x=e^6t x->无穷解一道大一极限题 lim(x→1)(1-x^2)/sinπxt=1-x,t-->0lim(2t-t^2)/sin(π-πt)=lim(2-t)t/sinπt=lim(2-t)t/πt=2/πlim(2-t)t/sinπt=lim(2-t)t/πt=2/π这部怎么来的设f(t)=lim(x→∞)t(1+2/x)^(x-t),求f'(t) lim→0[∫(上限x,下限0)(1+t^2)e^t^2dt]/xe^x^2