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I just can't solve this one (L'Hopital not allowed).
The limit of (2^x) / ln(x) when x goes to +infinity.
I'm trying to use e^x >> ln(x). So 2^x = e^(xln(2)). If u = xln(2) then we have the limit when u goes to +infinity of (e^u) / ln(u / ln(2)) = (e^u) / (ln(u) - ln(ln(2)).
So, I tried (e^u) / ln(u) < (e^u) / (ln(u) - ln(ln(2))) but THAT'S A LIE since ln(ln(2)) < 0, and that was my last hope.
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Isn't it just infinity?
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>>364250
it is, but I can't just say "it's just infinity"
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>>364243
You can use 2^x/ln(x) > 2^x/x for x>=1 (or another function) and prove that 2^x/x is unbounded.
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2^x/ln x => d/dx (2^x) / d/dx (ln x) => x2^(x-1) / 1/x
=> x^2 * 2^(x-1)
limit is +infinity
Also 2^x grows faster than ln x so inifnity too
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