高數(shù)入門(mén)求解

(4)

lim(x->1) (x^n-1)/(x-1)

=lim(x->1) [x^(n-1)+x^(n-2)+...+x+1]

=1+1+...+1

=n

(2)

f(x) =xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

case 1: x<-1

f(x)?

=xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

=xlim(n->∞) [1/x^(2n)-1)]/[1/x^(2n)+1 ]

=x(0-1)/(0+1)

=-x

case 2: x=-1

f(x) =xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

f(-1) = x( 1-1)/(1+1) =0

case 3: -1

f(x)?

=xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

=x(1-0)/(1+0)

=x

case 4: x=1

f(x) =xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

f(1) = x( 1-1)/(1+1) =0

case 5 : x>1

f(x)

=xlim(n->∞) [1-x^(2n)]/[1+x^(2n) ]

=xlim(n->∞) [1/x^(2n)-1)]/[1/x^(2n)+1 ]

=x(0-1)/(0+1)

=-x