Question

求极限值，

请手写，或发图

Answer 1

y->0

√( 1+2y) = 1+(1/2)(2y) -(1/8)(2y)^2 +o(y^2)  =1+y- (1/2)y^2 +o(y^2)

√(1+y) = 1+(1/2)y-(1/8)y^2 +o(y^2)

1+√( 1+2y) - 2√(1+y)

=1+[1+y- (1/2)y^2 +o(y^2)] -2[1+(1/2)y-(1/8)y^2 +o(y^2)]

=(-1/2 + 1/4)y^2 +o(y^2)

=-(1/4)y^2 +o(y^2)

lim(x->+∞) x.[ x+√(x^2+2x) - 2√(x^2+x) ]

y=1/x

=lim(y->0) [ (1/y)+√( 1/y^2+2/y) - 2√(1/y^2+1/y) ] /y

=lim(y->0) [ 1+√( 1+2y) - 2√(1+y) ] /y^2

=lim(y->0) (-1/4)y^2 /y^2

=-1/4

Answer 2

分子有理化，
分子分母同乘以x+√(x²+2x)+2√(x²+x)
=limx(2x√(x²+2x)-2x²-2x)/(x+√(x²+2x)+2√(x²+x))
=lim2x(√(x²+2x)-x-1)/(1+√(1+2/x)+2√(1+1/x))
=lim2x(-1)/(1+1+2)(√(x²+2x)+x+1)
=lim-1/2(√(1+2/x)+1+1/x)
=-1/4

追答

分子分母有理化，变减为加后整个因子化简，无穷大倒数为零性质学会应用