Question #10bb7

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Question #10bb7

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Answer 1 · 2017-12-17T23:29:04

$e^{- \frac{5}{2}}$ $e^(-5/2)$

Explanation:

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

${(\frac{4 n}{4 n + 5})}^{2 n}$ $((4n)/(4n + 5))^(2n)$ = ${(\frac{(4 n + 5) - 5}{4 n + 5})}^{2 n}$ $(((4n+5)-5)/(4n + 5))^(2n)$

$= {(1 - \frac{5}{4 n + 5})}^{2 n}$ $= (1 - 5/(4n + 5))^(2n)$

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

$= {(1 - \frac{5}{α + 5})}^{\frac{α}{2}}$ $= (1 - 5/(alpha + 5))^(alpha/2)$

$= e^{- \frac{5}{2}}$ $= e^(-5/2)$

Answer 2 · 2017-12-18T01:50:09

$e^{- \frac{5}{2}}$ $e^(-5/2)$

Explanation:

Making $y = 4 n + 5$ $y = 4n+5$ we have

${(\frac{y - 5}{y})}^{\frac{y - 5}{2}} = {(1 - \frac{5}{y})}^{\frac{y}{2}} {(1 - \frac{5}{y})}^{- \frac{5}{2}}$ $((y-5)/y)^((y-5)/2) = (1-5/y)^(y/2) (1-5/y)^(-5/2)$

now making $ξ = - \frac{y}{5}$ $xi = -y/5$ and substituting

${(1 - \frac{5}{y})}^{\frac{y}{2}} {(1 - \frac{5}{y})}^{- \frac{5}{2}} = {(1 + \frac{1}{ξ})}^{- \frac{5}{2} ξ} {(1 + \frac{1}{ξ})}^{- \frac{5}{2}} =$ $(1-5/y)^(y/2) (1-5/y)^(-5/2) = (1+1/xi)^(-5/2 xi)(1+1/xi)^(-5/2) =$

$= {({(1 + \frac{1}{ξ})}^{ξ})}^{- \frac{5}{2}} {(1 + \frac{1}{ξ})}^{- \frac{5}{2}}$ $= ( (1+1/xi)^xi)^(-5/2)(1+1/xi)^(-5/2)$

Now $n \to \infty \Rightarrow y \to \infty \Rightarrow ξ \to - \infty$ $n->oo rArr y->oo rArr xi -> -oo$ 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

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

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Question #10bb7

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