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

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### 3 Answers

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

The limit `lim_(x->oo)x^2/(x^2+1)` , has to be determined.

If we have two numbers x and y and x > y, 1/x < 1/y. This is true for any two numbers. If the value of a number approaches `oo` , its inverse is equal to 0. Or we can say if `x -> oo` , `1/x -> 0`

Now take the limit `lim_(x->oo)x^2/(x^2+1)`

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

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

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

As `x -> oo` , `1/(x^2 + 1) -> 0`

The limit is therefore equal to 1 - 0 = 1

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

According to the rule, if x approaches to `oo` , and the degree of the variable x from expression from numerator is equal to the degree of variable from expression from denominator, the limit is the ratio of leading coefficients:

lim x^2/(x^2 + 1) = 1/1 = 1

x->+`oo`

Of course, we can evaluate the limit using L'Hospital rule, but this way of solving seems much easier in this case.

**Therefore, the requested limit of the given function, if x approaches to + `oo` , is 1.**