lim(x趋近零)[∫(1+t^2) e^(t^2-x^2)d(x)]⼀x^2 {定积分上限是x^2,下限为0}

2025-01-21 15:33:49
推荐回答(2个)
回答1:

先整理分子,将带x的拿到积分外
∫(1+t^2) e^(t^2-x^2)d(x)=e^(-x^2)∫(1+t^2) e^(t^2)d(x)
然后用洛必达法则
原式=lim(x趋近零)e^(-x^2)∫(1+t^2) e^(t^2)d(x)/x^2
=lim(x趋近零)e^(-x^2)*lim∫(1+t^2) e^(t^2)d(x)/x^2
=1*lim(1+x^4) e^(x^4)/2x
=无穷大

回答2:

lim(x→0) [∫(0,x²) (1+t²)e^(t²-x²)dt] / x²
= lim(x→0) 2x(1+x²)e^(x²-x²) / (2x) <= 洛必达法则
= lim(x→0) (1+x²)e^(0)
= (1+0)(1)
= 1