I'm studying the function $f(x):=2^{1−1/x}−x^{−1/x}−1$ and want to show that there is a $n>0$ such that $f(x)>0 \, \; \forall x>n$. Do you have any suggestion? I tried to show that this function has a single maximum but I wasn't successful.
Behavior of $2^{1−1/x}−x^{−1/x}−1$
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calculus
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0I deleted my answer because the hint I gave was not helpful. – 2011-04-21
2 Answers
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Expanding the exponentials in terms of $1/x$ yields
$f(x)=2(1-\ln2/x)-(1-\ln x/x)-1+O\left(\left(\frac{\ln x}{x}\right)^2\right)=\frac{\ln x - 2\ln2}{x}+O\left(\left(\frac{\ln x}{x}\right)^2\right)\;.$
For sufficiently large $x$ the first term is positive and larger in magnitude than the $O((\ln x/x)^2)$ term.
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0Thanks for your reply! Now I understand it. Very good your answer! – 2011-04-21
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Consider the function $f(1/x)=2/2^x-x^x-1$. This function is continuous on $[0,\infty)$ and takes the value zero at $x=0$. Its derivative is $-2\log(2)/2^x-x^x(\log(x)+1)$ which must be positive near $x=0$, because $\log(x)$ goes to $-\infty$ as $x\downarrow 0$.
Thus $f(1/x)$ is positive for $x$ near zero, so $f(x)$ is positive for sufficiently large $x$.