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lim∫sintDt x

lim(x→0) [∫(0→x) sint dt]/x² = lim(x→0) [∫(0→x) sint dt'/(x²)' = lim(x→0) sinx/(2x) = 1/2 · lim(x→0) (sinx)/x = 1/2

x->0 sinx ~ x lim(x->0) (∫(0->x^2) sint dt)/ [x(sinx)^3] =lim(x->0) (∫(0->x^2) sint dt)/ x^4 (0/0) =lim(x->0)2xsin(x^2)/ (4x^3) =lim(x->0)2x^3/ (4x^3) =1/2

求采纳

如图所示:

由洛必达法则 原式 = lim(x→0) xsinx / [ 3x^2 / (1+x^3) ] = lim(x→0) (1+x^3)sinx / (3x) = 1/3

由于是0比0型,直接罗比达法则,分子求导为sinx,分母求导为2x,所以答案为1/2

根据洛比达法则,上下同时求导2次: 原式 =lim(x->0) arcsinx/2x =lim(x->0) [1/√(1-x^2)] / 2 =1/2

因为当x→0时,sinx~x,tanx~x,所以,limx→+0∫sinx0tantdt∫tanx0sintdt =limx→+0∫x0tdt∫x0tdt=1.故答案为:1.

解答:解; 由题知,f′(0)=limx→0f(x)-f(0)x-0≠0∴limx→0∫x20(x2-t)sintdtxk+1=limx→0x2∫x20sintdt-∫x20tsintdtxk+1=limx→02x∫x20sintdt(k+1)xk=limx→02∫x20sintdt+4x2sinx2(k+1)kxk-1=limx→012xsinx2+8x3sinx2(k+1)k(k-1)xk-2=limx→012x3+8x5(k+...

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