YLL討論網

由訪客於星期一八月 01, 2005 5:52 pm

(3)lim tan2x/3x = lim (2/3)*[(sin2x)/(2x)]*(1/cos2x) = (2/3)*1*1=2/3

(4)lim (3x^3-4x^2+2x-1)/(4x^3+5x^2-2x+6) = 3/4

由訪客於星期一八月 01, 2005 5:38 pm

(1)lim (sin7x)/(4x) = lim [(sin7x)/(7x)]*7/4 = 7/4 (x->0)

(2)lim [x-√(x^2-2x)] = lim (2x)/[x+√(x^2-2x)] = lim [2/(1+√(1-2/x)] = 1

由訪客於星期三七月 13, 2005 2:37 pm

(1) lim  sin7x/4x
    x->o

(2) lim  (x-根號x^2-2x)
    x->∞

(3) lim tan2x/3x
    x->o

(4) lim   (3x^3-4x^2+2x-1)/(4x^3+5x^2-2x+6)
  x->-∞

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[quote="Anonymous"](1)lim (sin7x)/(4x) = lim [(sin7x)/(7x)]*7/4 = 7/4 (x->0) (2)lim [x-√(x^2-2x)] = lim (2x)/[x+√(x^2-2x)] = lim [2/(1+√(1-2/x)] = 1[/quote]