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limsin2x sin3x

解法一:等价无穷小 lim sin3x/sin2x x→0 =lim 3x/(2x) x→0 =3/2 解法二:高中解法,运用两个重要极限的第一个重要极限 lim sin3x/sin2x x→0 =lim (3/2)[sin3x/(3x)]/[sin2x/(2x)] x→0 =(3/2)·1/1 =3/2

无穷小代换: 原式=lim (x-2x)/(x+3x) =lim -x/(4x) =-1/4

let y=π-x lim(x->π) sin2x/sin3x =lim(y->0) sin2y/sin3y =lim(y->0) 2y/(3y) =2/3 ans : A

lim(x→0)sin3x/sin2x =lim(x→0)3x(sin3x/3x)/[(sin2x/2x)/2x] =3/2

解: lim sin(2x)/sin(3x) x→0 =lim (2/3)[sin(2x)/(2x)]/[sin(3x)/(3x)] x→0 =lim (2/3)· (1/1) x→0 =2/3

x→ 0时,sin3x~3x,所以原式=2/3

原式=lim[ln(1+2x)/x]*lim(x/sin3x)=2*(1/3)=2/3

1. lim sin(2x)/ sin(3x) x→0 =lim (2x)/(3x) x→0 =⅔ 2. 令f(x)=sinx,则f'(x)=cosx lim (sinx-sina)/(x-a) x→a =f'(a) =cosa 3. lim (1- 3/x)^x x→∞ =lim [(1+(-3/x)]^(-x/3)]⁻³ x→∞ =e⁻³

=lim(sinx^3/x^3)•(x^3/sin^2x)=lim(x^3/sin^2x)=lim x•(sinx/x)^(-2)=0

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