Bruce Sherman

2022-10-09

Why does $\underset{x\to \mathrm{\infty}}{lim}x-{x}^{{\frac{1}{x}}^{\frac{1}{x}}}-{\mathrm{log}}^{2}x=0?$

Moreover, why is

$x-{x}^{{\frac{1}{x}}^{\frac{1}{x}}}\approx {\mathrm{log}}^{2}x?$

Moreover, why is

$x-{x}^{{\frac{1}{x}}^{\frac{1}{x}}}\approx {\mathrm{log}}^{2}x?$

Gabriella Hensley

Beginner2022-10-10Added 6 answers

By a Taylor expansion,

${\left(\frac{1}{x}\right)}^{1/x}=\mathrm{exp}(-\frac{1}{x}\mathrm{ln}x)=1-\frac{\mathrm{ln}x}{x}+O\left(\frac{{\mathrm{ln}}^{2}x}{{x}^{2}}\right).$

Thus the second term equals

$(1+\delta )\mathrm{exp}(\mathrm{ln}x(1-\frac{\mathrm{ln}x}{x}))=x(1-\frac{{\mathrm{ln}}^{2}x}{x}+O\left(\frac{{\mathrm{ln}}^{4}x}{{x}^{2}}\right))(1+\delta )$

with $\delta =O({\mathrm{ln}}^{3}x/{x}^{2})$, so this gives the claim.

${\left(\frac{1}{x}\right)}^{1/x}=\mathrm{exp}(-\frac{1}{x}\mathrm{ln}x)=1-\frac{\mathrm{ln}x}{x}+O\left(\frac{{\mathrm{ln}}^{2}x}{{x}^{2}}\right).$

Thus the second term equals

$(1+\delta )\mathrm{exp}(\mathrm{ln}x(1-\frac{\mathrm{ln}x}{x}))=x(1-\frac{{\mathrm{ln}}^{2}x}{x}+O\left(\frac{{\mathrm{ln}}^{4}x}{{x}^{2}}\right))(1+\delta )$

with $\delta =O({\mathrm{ln}}^{3}x/{x}^{2})$, so this gives the claim.

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