Question

y-->+无穷大,求lim(x^2+y^2)*e^-(x+y)

Answer 1

设x=rsint

y=rcost

因为x和y趋于正无穷,所以r充分大以后t必属于(0,pi/2)

0≤

|(x^2+y^2)e^-(x+y)|

=|r^2/e^(r(sint+cost))|

=|r^2/e^(√2*r*sin(t+pi/4))|

≤|r^2/e^r|→0

所以由迫敛性

lim （x^2+y^2）e^-(x+y) = 0

x->+∞

y->+∞

而且是绝对收敛

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