请用洛必达法则求导,

1个回答

  • lim(x->0) [e^x- e^(-x)]/sinx (0/0)

    =lim(x->0) [e^x+ e^(-x)]/cosx

    = 2

    lim(x->a) (x^m - a^m)/ (x^n - a^n )

    case 1:m>n

    lim(x->a) (x^m - a^m)/ (x^n - a^n ) (0/0)

    =lim(x->a) (mx^(m-1)/ (nx^(n-1)) (0/0)

    =..

    =...

    =lim(x->a) m(m-1)...(m-n+1)x^(m-n)/ n!

    = m(m-1)...(m-n+1).a^(m-n)/ n!

    = m!.a^(m-n)/[n!(m-n)!)]

    case 2:m=n

    lim(x->a) (x^m - a^m)/ (x^n - a^n ) =1

    case 3:ma) (x^m - a^m)/ (x^n - a^n ) (0/0)

    =lim(x->a) (mx^(m-1)/ (nx^(n-1)) (0/0)

    =..

    =...

    =lim(x->a) m!/[ n(n-1).(n-m+1).x^(n-m)]

    = m!(n-m)!/ [n!.a^(n-m)]

    lim(x->0)x^2.e^(1/x^2)

    =lim(x->0)e^(1/x^2) /(1/x^2) ( ∞/ ∞)

    =lim(x->0) (-2/x^3).e^(1/x^2) /(-2/x^3)

    =lim(x->0) e^(1/x^2)

    -> ∞