# What is limit of x*(pie/2-arctgx) when x go to `oo?`

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### 1 Answer

You need to evaluate the limit, hence, you need to replace `oo` for `x` in equation, such that:

`lim_(x->oo) x*(pi/2 - arctan x) = oo*(pi/2 - arctan oo)`

`lim_(x->oo) x*(pi/2 - arctan x) = oo*(pi/2 -pi/2)`

`lim_(x->oo) x*(pi/2 - arctan x) = oo*0`

Since the limit is indeterminate, you need to use l'Hospital's theorem, such that:

`lim_(x->oo) (pi/2 - arctan x)/(1/x) =lim_(x->oo) ((pi/2 - arctan x)')/((1/x)')`

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

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

The indetermination oo/oo requests for you use l'Hospital's theorem, such that:

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

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

**Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->oo) x*(pi/2 - arctan x) = 1` .**

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