Question

二重极限lim(x,y)→(0,0)(x+y)ln(x^2+y^2)

二重极限lim(x,y)→(0,0)(x+y)ln(x^2+y^2)

Answer 1

(x+y)²≤2(x²+y²)
0=lim-√(2u)lnu≤lim≤lim√(2u)lnu=0
夹逼准则，lim=0

Answer 2

x=rcosm y=rsinm
原式=
lim(r趋于0)2r(cosm+sinm)lnr

lim(r趋于0)rlnr(洛必达)=0
无穷小量×有界=0