What is lim In (n go to infinite) if In = integral (1 to 2) x^n/(x^n+1)?

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You need to evaluate the limit of the given definite integral `int_1^2 x^n/(x^n+1) dx` , hence, you need to evaluate the definite integral, such that:

`int_1^2 x^n/(x^n+1) dx = int_1^2 (x^n + 1 - 1)/(x^n+1) dx`

`int_1^2 x^n/(x^n+1) dx = int_1^2 (1 - 1/(x^n+1)) dx`

Using the proeprty of linearity of integral yields:

`int_1^2 x^n/(x^n+1) dx = int_1^2 dx - int_1^2 (1/(x^n+1)) dx`

You should evaluate the definite integral `int_1^2 (1/(x^n+1)) dx` using the following inequality, such that:

`int_1^2 (1/(x^n+1)) dx < int_1^2 (1/x^n) dx`

`int_1^2 (1/x^n) dx = int_1^2 x^(-n) dx = x^(1-n)/(1-n)|_1^2`

`int_1^2 (1/x^n) dx = (2^(1-n) - 1)/(1 - n)`

`int_1^2 x^n/(x^n+1) dx = x|_1^2 - (2^(1-n) - 1)/(1 - n)`

`int_1^2 x^n/(x^n+1) dx = 1 - (2^(1-n) - 1)/(1 - n)`

You need to evaluate the limit, such that:

`lim_(n->oo)I_n = lim_(n->oo) (1 - (2^(1-n) - 1)/(1 - n)`

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

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

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

Hence, evaluating the limit of the given integral, yields `lim_(n->oo) int_1^2 x^n/(x^n+1) dx = 1` .

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