# What is domain of continuaty of y=lim ((x^2+1)^n+x-2)/(2(x^2+1)^n+x^2+2) , n go to infinte?

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### 1 Answer

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}` .**