# Calculate limit I(subscript n), with n -> `oo` I(subscript n)= `int_0^1` x^n/(x^2+x+1) dx

sciencesolve | Certified Educator

You need to evaluate the following limit, such that:

`lim_(n->oo) I_n = lim_(n->oo) int_0^1 (x^n)/(x^2 + x + 1) dx`

You may use the squeeze theorem to evaluate the limit, such that:

`f(x)<= h(x) <= g(x)`

If `lim_(x->x_0) f(x) = lim_(x->x_0) g(x) = l` , then `lim_(x->x_0) h(x) = l.`

Reasoning by analogy an starting from the fact that `I_n =int_0^1 (x^n)/(x^2 + x + 1) dx >= 0` , you need to find a function `g(x) >= h(x) = (x^n)/(x^2 + x + 1).`

Since `x^n > (x^n)/(x^2 + x + 1)` for `x in [0,1]` , yields:

`0 <= I_n <= int_0^1 x^n dx`

Evaluating the definite integral `int_0^1 x^n dx ` yields:

`int_0^1 x^n dx = x^(n+1)/(n+1)|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 x^n dx = 1^(n+1)/(n+1) - 0^(n+1)/(n+1)`

`int_0^1 x^n dx = 1/(n+1)`

Evaluating the limits yields:

`lim_(n->oo) 0 <= lim_(n->oo) I_n <= lim_(n->oo) 1/(n+1)`

`0 <= lim_(n->oo) I_n <= 1/oo -> 0 => lim_(n->oo) I_n = 0`

Hence, evaluating the given limit, using the squeeze theorem, yields `lim_(n->oo) I_n = 0.`