Which is the result for the limit in the image ?

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nick-teal | High School Teacher | (Level 3) Adjunct Educator

Posted on

After reviewing science solve's answer, the only issue his has is that there should not be a pi/2 left over after taking the derivative of the numerator. (during the l'Hospital part)

This means that line should read 

`ln l = lim_(n->oo) 1/(arctan n)*1/(1+n^2)/(-1/(n^2))`

Then this makes the last line read

`ln l = -2/pi => l = e^(-2/pi)`

or, `l = 1/e^(2/pi)`

Hope this helps! This was a crazy problem.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The evaluation of limit must be performed following the next steps:

`lim_(n->oo)(2/(pi)arctan n)^n = (2/(pi) lim_(n->oo) arctan n)^(lim_(n->oo) n)`

Evaluating `lim_(n->oo) arctan n` yields:

`lim_(n->oo) arctan n = arctan oo = (pi/2)`

- the next step is to replace `pi/2` instead of `lim_(n->oo) arctan n` , such that:

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

- Since the case `1^oo` is indeterminate, you need to create the special limit, such that:

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

- Hence, you need to take logarithm of limit, such that:

`l = lim_(n->oo)(2/(pi)arctan n)^n`

`ln l = ln lim_(n->oo) (2/(pi)arctan n)^n`

`ln l = lim_(n->oo) ln (2/(pi)arctan n)^n`

`ln l = lim_(n->oo) n*ln (2/(pi)arctan n)`

`ln l = lim_(n->oo) (ln (2/(pi)arctan n))/(1/n)`

`ln l = (ln (2/(pi) lim_(n->oo) arctan n))/(lim_(n->oo)(1/n))`

`ln l = (ln (2/(pi)* (pi/2)))/(lim_(n->oo)(1/n))`

`ln l = (ln 1)/(1/oo)`

`ln l = 0/0`

Since `0/0` is indeterminate, you need to use l'Hospital theorem:

`lim_(n->oo) (u(n))/(v(n)) = lim_(n->oo) (u'(n))/(v'(n))`

`ln l = lim_(n->oo) (pi)/2*1/(arctan n)*1/(1+n^2)/(-1/(n^2))`

`ln l = -(pi)/2*lim_(n->oo) (n^2)/(1+n^2)*lim_(n->oo) 1/(arctan n)`

You need to evaluating each limit, such that:

`lim_(n->oo) 1/(arctan n) = 1/((pi)/2)`

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

You need to reduce by `n^2` , such that:

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

Since ` lim_(n->oo) 1/(n^2) = 1/(oo) = 0`

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

Replacing the limit values:

`ln l = -(pi)/2* 2/(pi)*1 => ln l = -1 => l = e^(-1) => l = 1/e`

Hence, evaluating the limit, yields ` lim_(n->oo)(2/(pi)arctan n)^n = 1/e.`

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