Question

求以下求极限问题

Answer 1

方法如下，
请作参考：

Answer 2

解答如下 

Answer 3

(1)
lim(x->0) (1+3x^4)^[1/(sinx)^2]
=lim(x->0) (1+3x^4)^(1/x^2)
=1
(2)
lim(x->无穷) (3x^2+8x+1)^4/(3x^4+5x+2)^2
分子分母同时除x^8
=lim(x->无穷) (3+8/x+1/x^2)^4/(3+5/x^3+2/x^4)^2
=3^4/3^2
=9
(3)
lim(x->0) [√(1+sin(x^2)) -1 ]/[ e^(x^2) -1 ]
=lim(x->0) (1/2)x^2/ x^2
=1/2
(4)
y->0+
√(y^4+1) = 1+(1/2)y^4 +o(y^4)
3+√(y^4+1) = 4+(1/2)y^4 +o(y^4)
√【3+√(y^4+1)】
=√[ 4+(1/2)y^4 +o(y^4)]
=2.√[ 1+(1/8)y^4 +o(y^4)]
=2( 1+ (1/16)y^4) +o(y^4)
√【3+√(y^4+1)】 -2 =(1/8)y^4 +o(y^4)
//
lim(x->+无穷) 2x^2. [ √【3x^4+x^2.√(x^4+1)】 -√(4x^4) ]
y=1/x
=lim(y->0+) (2/y^2). [ √【3/y^4+(1/y^2).√(1/y^4+1)】 -√(4/y^4) ]
=lim(y->0+) 2 [ √【3+√(y^4+1)】 -√4 ]/y^4
=2lim(y->0+) [ √【3+√(y^4+1)】 -2 ]/y^4
=2lim(y->0+) [ (1/8)y^4 ]/y^4
=1/4