<Method 1>
1. Prove：e^x/x → inf 當x→inf
Since e^x = 1+x+x^2/2!+...
so e^x/x = 1/x + 1 + x/2! +.... >= x/2! 當 x>0
so e^x/x → inf by trivial armugemt
2. x-lnx = ln(e^x/x)
Since lnx → inf 當 x→inf
e^x/x → inf 當 x→inf (from 1.)
so ln(e^x/x) = x - lnx → inf 當x→inf (見lemma)
<lemma>：f(x)→+inf , g(x)→+inf as x→inf
then f(g(x)) → +inf as x→inf
<pf>：We have f(x) >= M as x>= x_M , g(x) >= N as x>= x_N
take N = x_M
then g(x) >= x_M as x >= x_(x_M) =: x_M'
so f(g(x)) >= M (Since g(x) >= x_M) , as x>=M'
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<Method 2>
f(x) =: x - lnx ： [1,+inf) → R
f'(x) = 1 - 1/x > 0 as x > 1
so f(x) ↑ as x > 1 and f(x) >= f(1) = 1
Assume f is conv.
so f is bdd. by M > 1 (if M<= 1 , 與絕對遞增牴觸 )
so x - lnx = ln(e^x/x) <= M
=> e^x/x <= e^M 與e^x/x無界牴觸(Method 1的第一步)
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