You need to notice that`x^2 + 1 > 0`  for `x in R`   and `x^2 + 1 = 1`  if `x = 0` , hence, you may evaluate `f(0)`  such that:

`f(0) = lim_(n->oo) (1^n - 2)/(2*1^n + 2) = (1 - 2)/4 = -1/4`

You need to evaluate `f(x)`  such that:

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

You need to force the factor `(x^2+1)^n`  such that:

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

Reducing duplicate factors yields:

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

Since `lim_(n->oo)(x - 2)/((x^2+1)^n) = 0`  and `lim_(n->oo) (x^2+2)/((x^2+1)^n) = 0`  yields:

`f(x) = (1 + 0)/(2 + 0) = 1/2`

Hence, evaluating `f(x)`  yields `f(x) = {(1/2, x in R-{0}),(-1/4, x = 0):}, ` hence, the function is continuous over `R - {0}` .