这道题目怎么做的额

设f'(x) 连续,f(0)=0,f'(0)不等于0,求lim∫f(t)dt/∫f(t)dt
注明x趋向于0,前面的上限是x2,下限为0,后面的上限为x,下限为0,怎么求的?把过程写清楚点可以吗?我知道用洛必达法则但是不知道怎么用?

还有一到lim(e^x-1-x)/[(1-x)^1/2-cos(x^1/2)] x趋向于0 又是怎么做的额?帮帮忙谢谢

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lim_{x->0}∫_{0}^{x^2}f(t)dt/∫_{0}^{x}f(t)dt

= lim_{x->0}[2xf(x^2)]/[f(x)]

= 2lim_{x->0}[f(x^2) + 2x^2f'(x^2)]/[f'(x)]

= 2[f(0) +0f'(0)]/f'(0)

= 0.

lim_{x->0}(e^x-1-x)/[(1-x)^1/2-cos(x^1/2)]

= lim_{x->0}(e^x-1)/[-(1-x)^(-1/2)/2+sin(x^1/2)x^(-1/2)/2]

= 2lim_{x->0}(e^x-1)/[-(1-x)^(-1/2)+sin(x^1/2)x^(-1/2)]

= 2lim_{x->0}[(e^x-1)/x] [x(1-x)^(1/2)x^(1/2)]/[-x^(1/2) + sin(x^1/2)(1-x)^(1/2)]

= 2lim_{x->0}[x(1-x)^(1/2)x^(1/2)]/[-x^(1/2) + sin(x^1/悔滑2)(1-x)^(1/2)]

= 2lim_{x->0}[x^(3/2)]/[-x^(1/2) + sin(x^1/2)(1-x)^(1/2)]

{u = x^(1/2)}

= 2lim_{u->0}[u^3]/[-u + sin(u)(1-u^2)^(1/2)]

= 2lim_{u->0}[3u^2]/[-1 + cos(u)(1-u^2)^(1/2) - usin(u)(1-u^2)^(-1/蚂前竖2)]

= 6lim_{u->0}[u^2](1-u^2)^(1/2)/闷大[-(1-u^2)^(1/2) + cos(u)(1-u^2) - usin(u)]

= 6lim_{u->0}[u^2]/[-(1-u^2)^(1/2) + cos(u)(1-u^2) - usin(u)]

= 6lim_{u->0}[2u]/[u(1-u^2)^(-1/2) - sin(u)(1-u^2) - 2ucos(u) - sin(u) - ucos(u)]

= 6lim_{u->0}[2]/[(1-u^2)^(-1/2) - sin(u)(1-u^2)/u - 2cos(u) - sin(u)/u - cos(u)]

= 12/[1 - 1 - 2 - 1 - 1]

= -3

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