Question #10bb7

2 Answers
Dec 17, 2017

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)

Dec 18, 2017

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