lim(x趋于无穷大)(x^2+1/x+1 -ax-b)=0求a b

1个回答

  • lim(x->∞)(x^2+1)/(x+1) -ax-b = 0

    0=lim(x->∞)[(x^2+1)/(x+1) -ax-b ] / x

    =lim(x->∞)[(x+1/x)/(x+1) -a-b/x

    =lim(x->∞)[(1+1/x^2)/(1+1/x) -a-b/x

    = 1 - a - 0

    = 1-a

    ∴ a = 1

    lim(x->∞)(x^2+1)/(x+1) - x - b

    =lim(x->∞)(x^2+1 - x^2 - x )/(x+1) - b

    =lim(x->∞)(1 - x )/(x+1) - b

    =lim(x->∞)(1/x - 1 )/(1+1/x) - b

    = -1 - b

    = 0

    ∴ b = -1