# Behavior of $2^{1−1/x}−x^{−1/x}−1$

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.

• I deleted my answer because the hint I gave was not helpful. – Américo Tavares Apr 21 '11 at 8:44

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.

• @joriqui: Contrary to OP I do not follow your expansion, mainly because of the $x^{-1/x}$ term. Of course the fault is mine. Would you mind explaining the expansion of that $x^{-1/x}$ term? – Américo Tavares Apr 21 '11 at 13:32
• @Américo: I think the fault was in fact mine :-) There was a sign error. Is it clearer now? $x^{-1/x}=\mathrm e^{\ln x(-1/x)}=1-\ln x/x+O((\ln x/x)^2)$. – joriki Apr 21 '11 at 13:39
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 $x$.