Question

高等数学:利用无穷小等价代换求下列极限





Answer 1

(2)

x->0

分子

1-(cosx)^2 = (sinx)^2 = x^2 +o(x^2)

5x^2-2(1-(cosx)^2) = 3x^2+o(x^2)

分母

(tanx)^2 =x^2+o(x^2)

3x^3+4(tanx)^2 =4x^2+o(x^2)

lim(x->0) [5x^2-2(1-(cosx)^2)]/[3x^3+4(tanx)^2]

=lim(x->0) 3x^2/[4x^2]

=3/4

(5)

x->0

分子

√(1+x^2)  = 1+(1/2)x^2 +o(x^2)

√(1+x^2) -1  = (1/2)x^2 +o(x^2)

(1+tanx)^(1/3) = (1+x)^(1/3) = 1+(1/3)x+o(x)

(1+tanx)^(1/3) -1 = (1/3)x+o(x)

[(1+tanx)^(1/3) -1].[√(1+x^2) -1] = (1/6)x^3 +o(x^3)

分母

tanx = x+(1/3)x^3 +o(x^3)

sinx = x-(1/6)x^3 +o(x^3)

tanx -sinx =(1/2)x^3+o(x^3)

lim(x->0) [(1+tanx)^(1/3)-1] .[ √(1+x^2) -1 ] /(tanx -sinx)

=lim(x->0) (1/6)x^3 /[(1/2)x^3]

=1/3

(6)

x->0-

分子

cosx = 1-(1/2)x^2 +o(x^2)

√cosx = √[1-(1/2)x^2 +o(x^2)] = 1- (1/4)x^2 +o(x^2)

1-√cosx =  (1/4)x^2 +o(x^2)

tanx = x+o(x)

(1-√cosx).tanx = (1/4)x^3 +o(x^3)

分母

1-cosx = (1/2)x^2 +o(x^2)

(1-cosx)^(3/2) = (1/2^(3/2)) x^3 +o(x^3)

lim(x->0-) (1-√cosx).tanx/(1-cosx)^(3/2)

=lim(x->0-) (1/4)x^3 /[  (1/2^(3/2)) x^3  ]

=2^(-2 +3/2)

=2^(-1/2)

=√2/2

Answer 2

(5)原式-->[(1/3)tanx*x^2/2]/{tanx(1-cosx)}
-->(1/6)x^2/(x^2/2)
-->1/3.
1-cosx=2[sin(x/2)]^2等价于x^2/2,
tanx等价于x.