Calculate the limit of the function y=x^2/(x^2+1), if x --> +infinte?

Since ` ` ` ` `lim_(x->oo) ``x^2` = `oo` and `
lim_(x->oo) ` ` ` `x^2` +1 = ` `

We can use
L'Hopital's rule that ` `` ``lim_(x->oo)` `(f(x))/(g(x))` = `lim_(x->oo)`
`(f'(x))/(g'(x))`
if `lim_(x->oo)` f(x) = `oo` and
`lim_(x->oo)` g(x) = `oo`

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

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

We could also use the following analysis.  Using the fact
that

`lim_(x->oo)` 1/x^n = 0, and dividing both the
denominator and numerator by
x^2 (x^2/x^2 = 1 and (x^2+1)/x^2= 1 + 1/x^2) we
get

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

and again since
`lim_(x->oo)` 1/x^2 = 0 we get

`lim_(x->oo)`
(x^2)/(x^2+1) = `lim_(x->oo)` (1/(1+0)) = 1/1 =
1

This second method also shows
that

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