How calculate limits with x to infinite sign lim (integral 0-x t^n sq root(1+t^2) dt)/(x^(n+2))?



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You need to evaluate the following limit, such that:

`lim_(x->oo) (int_0^x t^n sqrt(1 + t^2) dt)/(x^(n+2)) = oo/oo`

The indetermination "`oo/oo` " allows you to use l'Hospital's theorem, such that:

`lim_(x->oo) (int_0^x t^n sqrt(1 + t^2) dt)/(x^(n+2)) = lim_(x->oo) ((int_0^x t^n sqrt(1 + t^2) dt)')/((x^(n+2))') `

`lim_(x->oo) ((int_0^x t^n sqrt(1 + t^2) dt)')/((x^(n+2))')= lim_(x->oo) (x^n sqrt(1 + x^2))/((n+2)x^(n+1))`

Factoring out `x^2` under the square root yields:

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

`lim_(x->oo) (x^n |x|sqrt(1/x^2 + 1))/((n+2)x^(n+1))`

Since `x->oo` , hence `|x| = x` , such that:

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

Reducing duplicate factors yields:

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

Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->oo) (int_0^x t^n sqrt(1 + t^2) dt)/(x^(n+2)) = 1/(n+2).`

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