求解:定积分+求极限 急

如题所述

lim(x→0)[∫(0→x)e^(t^2)dt]^2/∫(0→2x)sin2tdt
=lim(x→0)2∫(0→x)e^(t^2)dt*e^(x^2)/[sin(2*2x)*2](洛必达法则)
=2*lim(x→0)∫(0→x)e^(t^2)dt*e^(x^2)/(2sin4x)
=2*lim(x→0)[e^(x^2)*e^(x^2)+∫(0→x)e^(t^2)dt*e^(x^2)*2x]/(2*4cos4x)(洛必达法则)
=2*(1*1+0)/(2*4*1)
=1/2
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