利用积分上限函数的性质求极限lim(x->0) ∫(x,0) (cost^2 dt)/x

Question

如题所述

Answer 1

解：∵x²y²≤(x²+y²)²/4
∴0≤(x²+y²)^(x²y²)≤(x²+y²)^[(x²+y²)²/4]
∵lim(x->0,y->0){(x²+y²)^[(x²+y²)²/4]}=lim(t->0)[t^(t²/4)]
(令t=x²+y²)
=lim(t->0)[e^(t²lnt/4)]
(应用对数性质)
=e^[lim(t->0)(t²lnt/4)]
(应用初等函数的连续性)
=e^{lim(t->0)[lnt/(4/t²)]}
=e^{lim(t->0)[(1/t)/(-8/t³)]}
(∞/∞型极限，应用罗比达法则)
=e^{lim(t->0)[t²/(-8)]}
=e^0
=1
∴由两边夹定理知，lim(x->0,y->0)[(x²+y²)^(x²y²)]=1。