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

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