Question

How to resolve??

`limx->oo(2/piarctanx)^x=`
`=e^(limx->ooxln(2/piarctanx))`
`limx->ooln(2/piarctanx)/(1/x)`
`[0/0]=>` de l'Hospital
`limx->oo(2/pi*1/(x^2+1)*(1/((2/pi)arctanx)))/(-1/x^2)=-x^2/((x^2+1)arctanx)=[oo/oo]`
Is everything ok? Again l'Hospital? Does it make any sense?

Answer 1

The limit gives `e^(-2/pi) ~~ 0.529`

Explanation:

Start by calling the limit `L`.

`L = lim_(x-> oo) (2/piarctanx)^x`

`lnL = lim_(x->oo) xln(2/piarctanx)`

`lnL = lim_(x-> oo) ln(2/pi arctanx)/(1/x)`

Now is the time to use l'hospitals

`lnL =lim_(n-> oo) (1/(2/piarctanx) * 2/pi 1/(1 + x^2))/(-1/x^2)`

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

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

L'hopitals again!!!

`lnL = lim_(n ->oo) (-2x)/(2xarctanx)`

`lnL = lim_(n-> oo) -1/(arctanx)`

Now evaluate through substitution.

`lnL = -1/(pi/2)`

`L = e^(-2/pi) ~~ 0.529`

Hopefully this helps!

Answer 2

`lim_(x rarr oo) (2/pi arctanx)^(x) = e^(-2/pi)`

Explanation:

We seek:

`L = lim_(x rarr oo) (2/pi arctanx)^(x)`

Put `x=1/u iff u=1/x => u rarr 0 " as " x rarr oo`, so we can write

`L = lim_(u rarr 0) (2/pi arctan(1/u))^(1/u)`

We can take Natural logarithms:

`ln L = ln{lim_(u rarr 0) (2/pi arctan(1/u))^(1/u)}`

Using the monotonicity of the logarithmic function we can write:

`ln L = lim_(u rarr 0) {ln(2/pi arctan(1/u))^(1/u)}`

Then using the properties of logarithms:

`ln L = lim_(u rarr 0) {1/u ln(2/pi arctan (1/u))}`
`\ \ \ \ \ \ = lim_(u rarr 0) {(ln(2/pi arctan (1/u)))/u}`

Noting that as:

`u rarr 0=> 2/pi arctan (1/u) rarr 2/pi * pi/2 =1` and `ln1=0`

Then we have an indeterminate form `0/0`, so we can apply L'Hôpital's rule to get:

`ln L = lim_(u rarr 0) {(d/(du) ln(2/pi arctan (1/u)))/(d/(du)u)}`
`\ \ \ \ \ \ = lim_(u rarr 0) {( 1/(2/pi arctan(1/u)) * (2/pi)/(1+(1/u)^2) * (-1/u^2) )/(1)}`

`\ \ \ \ \ \ = lim_(u rarr 0) {1/arctan(1/u) 1/(1+1/u^2) ( -1/u^2) }`

`\ \ \ \ \ \ = lim_(u rarr 0) {- 1/arctan(1/u) 1/(1+u^2) }`

And we can now readily evaluate this limit, thus:

`ln L = - 1/(pi/2) 1/(1+0)`

`\ \ \ \ \ \ = - 2/pi`

And so:

`L = e^(-2/pi)`