# What is limit of x*arctg(1/x) if x go to `oo`

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### 1 Answer

You need to evaluate the given limit, such that:

`lim_(x->oo) x*arctan(1/x) = oo*arctan(1/oo) = oo*0`

You may convert the indetermination oo*0 into 0/0, such that:

`lim_(x->oo) x*arctan(1/x) = lim_(x->oo) (arctan(1/x))/(1/x) = 0/0`

The indetermination case `0/0` requests for you to use l'Hospital's theorem, such that:

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

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

Reducing the duplicate factors, yields:

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

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

Using again l'Hospital's theorem yields:

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

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

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