Question

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$lim_{x to +infty} frac{ln(1+x^{1/2}+x^{1/3})}{ln(1+x^{1/3}+x^{1/4})}$

Answer 1

$large displaystylelim_{x to +infty}frac{lnleft(1+sqrt{x}+sqrt[3]{x}right)}{lnleft(1+sqrt[3]{x}+sqrt[4]{x}right)}=lim_{x to +infty}frac{lnleft(sqrt{x}left(frac{1}{sqrt{x}}+1+frac{1}{sqrt[6]{x}}right)right)}{lnleft(sqrt[3]{x}left(frac{1}{sqrt[3]{x}}+1+frac{1}{sqrt[12]{x}}right)right)}=\ \ \ =lim_{x to +infty}frac{lnsqrt{x}}{lnsqrt[3]{x}}=lim_{x to +infty}frac{1/2ln x}{1/3ln x}=frac{3}{2}$