use hospital calculate of limit 1/x^2 integralbetween x^2+1 and x^4+1 of sint/t x goes in 0

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to consider the function F(x). If F(x) is differentiated yields `f(x) = sin x/x` , hence `int_(x^2+1)^(x^4+1) (sint dt)/t = F(x^4 + 1) - F(x^2+ 1` ).

You need to evaluate the limit using l'Hospital's rule such that:

`lim_(x-gt0) (1/x^2)int_(x^2+1)^(x^4+1) (sint dt)/t = lim_(x-gt0) (F(x^4 + 1) - F(x^2+ 1))/(x^2)`

Plugging x = 0 in `(F(x^4 + 1) - F(x^2+ 1))/(x^2)`  yields:

`(F(0^4 + 1) - F(0^2+ 1))/(0^2) = 0/0`

This indetermination needs l'Hospital's rule such that:

`lim_(x-gt0) (F(x^4 + 1) - F(x^2+ 1))/(x^2) = lim_(x-gt0) (F'(x^4 + 1)(x^4 + 1)' - F'(x^2+ 1)(x^2 + 1)')/((x^2)')`

`lim_(x-gt0) (F(x^4 + 1) - F(x^2+ 1))/(x^2) = lim_(x-gt0) (4x^3*F'(x^4 + 1) - 2x*F'(x^2+ 1))/(2x)`

Reducing by 2x yields:

`lim_(x-gt0) (F(x^4 + 1) - F(x^2+ 1))/(x^2) = lim_(x-gt0) (2x^2*F'(x^4 + 1) - F'(x^2+ 1))`

You need to remember that `F'(x) = f(x)`  such that:

`F'(x^4 + 1) = (sin(x^4 + 1))/(x^4 + 1)`

`` `F'(x^2 + 1) = (sin(x^2 + 1))/(x^2 + 1)`

You need to write the new form of limit such that:

`lim_(x-gt0) (2x^2*(sin(x^4 + 1))/(x^4 + 1) - (sin(x^2 + 1))/(x^2 + 1))= lim_(x-gt0) (2x^2*(sin(x^4 + 1))/(x^4 + 1)) - lim_(x-gt0)(sin(x^2 + 1))/(x^2 + 1))`

You need to remember that `lim_(x-gt0) sin x/x = 1`

`lim_(x-gt0) (2x^2*(sin(x^4 + 1))/(x^4 + 1) - (sin(x^2 + 1))/(x^2 + 1))=2*0^2*1 - 1 = -1`

Hence, evaluating the limit yields `lim_(x-gt0) (1/x^2)int_(x^2+1)^(x^4+1) (sint dt)/t = -1.`