Question

Question #f1ea6

Answer 1

`lim_(x->1) (x^(n+1)-(n+1)x+n)/(x-1)^2 =(n(n+1))/2`

Explanation:

Simplify the function:

`(x^(n+1)-(n+1)x+n)/(x-1)^2 = (x^(n+1)-x -nx+n)/(x-1)^2`

`(x^(n+1)-(n+1)x+n)/(x-1)^2 = (x(x^n-1) -n(x-1))/(x-1)^2`

`(x^(n+1)-(n+1)x+n)/(x-1)^2 = [x/(x-1) (x^n-1)/(x-1)] -n/(x-1)`

Note that:

`(x^n-1)/(x-1) = 1+x+x^2+...+x^(n-1)`

so:

`(x^(n+1)-(n+1)x+n)/(x-1)^2 = [x/(x-1) ( 1+x+x^2+...+x^(n-1))] -n/(x-1)`

and finally:

`(1) " " (x^(n+1)-(n+1)x+n)/(x-1)^2 =( x+x^2+...+x^n -n)/(x-1)`

Consider now the function:

`f(x) = x+x^2+...+x^n`

Clearly `f(1) = n`, thus:

`lim_(x->1) (x^(n+1)-(n+1)x+n)/(x-1)^2 = lim_(x->1) ( x+x^2+...+x^n -n)/(x-1) = lim_(x->1) (f(x) -f(1))/(x-1)`

and by the definition of derivative:

`lim_(x->1) (x^(n+1)-(n+1)x+n)/(x-1)^2 = f'(1)`

As:

`f'(x) = 1+2x+...+nx^(n-1)`

we have:

`f'(1) = 1+2+...+n = (n(n+1))/2`

so:

`lim_(x->1) (x^(n+1)-(n+1)x+n)/(x-1)^2 =(n(n+1))/2`

Answer 2

We seek:

`lim_(x rarr 1)(x^(n+1)-(n+1)x+n)/(x-1)^2 = 1/2 n(n+1)`

Explanation:

We seek:

`L = lim_(x rarr 1)(x^(n+1)-(n+1)x+n)/(x-1)^2`

Both the numerator and the denominator `rarr 0` as `x rarr 0`. thus the limit `L` (if it exists) is of an indeterminate form `0/0`, and consequently, we can apply L'Hôpital's rule to get:

`L = lim_(x rarr 1) ( d/dx( x^(n+1)-(n+1)x+n) ) / ( d/dx (x-1)^2 )`

`\ \ = lim_(x rarr 1) ( (n+1)x^n - (n+1) ) / ( 2(x-1) )`

Again, we have an indeterminate form, so we can apply L'Hôpital's rule again:

`L = lim_(x rarr 1) ( d/dx( (n+1)x^n - (n+1) ) ) / ( d/dx( 2(x-1) ) )`

`\ \ = lim_(x rarr 1) ( n(n+1)x^(n-1) ) / ( 2x )`

And we can now directly evaluate this limit:

`L = ( n(n+1) xx 1^(n-1) ) / ( 2xx1 )`

`\ \ = ( n(n+1) ) / ( 2 )`