Question

Question #10bb7

Answer 1

`e^(-5/2)`

Explanation:

We will use the fact that `e` can also be expressed at the limit as n approaches infinity of `(1 + 1/n)^n`. Our goal now will be to express our expression in a similar form as e.

`((4n)/(4n + 5))^(2n)` = `(((4n+5)-5)/(4n + 5))^(2n)`

`= (1 - 5/(4n + 5))^(2n)`

If we now set `alpha` equal to `4n`, we will have

`= (1 - 5/(alpha + 5))^(alpha/2)`

`= e^(-5/2)`

Answer 2

`e^(-5/2)`

Explanation:

Making `y = 4n+5` we have

`((y-5)/y)^((y-5)/2) = (1-5/y)^(y/2) (1-5/y)^(-5/2)`

now making `xi = -y/5` and substituting

`(1-5/y)^(y/2) (1-5/y)^(-5/2) = (1+1/xi)^(-5/2 xi)(1+1/xi)^(-5/2) =`

`= ( (1+1/xi)^xi)^(-5/2)(1+1/xi)^(-5/2)`

Now `n->oo rArr y->oo rArr xi -> -oo` and

`lim_(n->oo)((4 n)/(4 n + 5))^(2 n) = lim_(xi->-oo)( (1+1/xi)^xi)^(-5/2)(1+1/xi)^(-5/2) =`

`(lim_(xi->-oo) (1+1/xi)^xi)^(-5/2)lim_(xi->oo)(1+1/xi)^(-5/2) = e^(-5/2)`

NOTE

We were using the fact

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