# How calculate limit lim f(subscript n)(x)/(x^(n+2)), x--->oo, with f(subscript n)(x) = integral (0 to x) t^n sqroot(t^2+1)dt?

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You need to evaluate the given limit such that:

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

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

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

Taking out the constant `1/(n+2)` and reducing duplicate factors `x^n` yields:

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

You need to force factor `x^2` under the square root, such that:

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

Since `sqrt x^2 = |x|` yields:

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

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

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

Reducing duplicate factors, yields:

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

Using the property of limit yields that the limit of the square root is the square root of the limit, such that:

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

Since `lim_(x->oo) 1/x^2 = 0` yields:

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

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

**Hence, evaluating the given limit, under the given conditions, yields `lim_(x->oo) (f_n(x))/(x^(n + 2)) = 1/(n+2)` .**