calculate limit n (n+1)S (1/(n+1)_1/n )arctg n x dx ?

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You should consider the function `f(x) = tan^(-1)(nx)` , that is continuous, hence, you may use mean value theorem over the interval `[1/n,1/(n+1)]` , such that:

`EE` c in [1/n,1/(n+1)] such that:

`f(c)*(1/n - 1/(n+1)) = int_(1/(n+1))^(1/n) tan^(-1)(nx) dx`

Replacing `f(c)*(1/n - 1/(n+1))` for `int_(1/(n+1))^(1/n) tan^(-1)(nx) dx` in limit, yields:

`lim_(n->oo) n(n+1)int_(1/(n+1))^(1/n) tan^(-1)(nx) dx = lim_(n->oo) n(n+1) f(c)*(1/n - 1/(n+1))`

`lim_(n->oo) n(n+1) f(c)*(1/n - 1/(n+1)) = lim_(n->oo) n(n+1) f(c)*(n+1-n)/(n(n+1))`

`lim_(n->oo) n(n+1) f(c)*(1/n - 1/(n+1)) = lim_(n->oo) n(n+1) f(c)*1/(n(n+1))`

Reducing duplicate factors yields:

`lim_(n->oo) n(n+1) f(c)*(1/n - 1/(n+1)) = lim_(n->oo) f(c) `

`lim_(n->oo) f(c)= lim_(n->oo) tan^(-1)(nc)`

Using squeeze theorem yields:

`lim_(n->oo) tan^(-1)(nc) < lim_(n->oo) (nc) = 1`

`lim_(n->oo) tan^(-1)(nc) = tan^(-1) 1 = pi/4`

**Hence, evaluating the limit, using mean value theorem and squeeze theorem, yields **`lim_(n->oo) n(n+1)int_(1/(n+1))^(1/n) tan^(-1)(nx) dx = pi/4.`

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