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My question is:

Show the limit of

[tex]x_{n}=\frac{ln(1+\sqrt{n}+\sqrt[3]{n})}{ln(1+\sqrt[3]{n}+\sqrt[4]{n})}[/tex]

as n approaches infinity

Solution:

[tex]{x_n} = \frac{{\ln (1 + {n^{\frac{1}{2}}} + {n^{\frac{1}{3}}})}}{{\ln (1 + {n^{\frac{1}{3}}} + {n^{\frac{1}{4}}})}} = \frac{{\ln \left( {{n^{\frac{1}{2}}} \cdot (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})} \right)}}{{\ln \left( {{n^{\frac{1}{3}}} \cdot (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{\frac{1}{{12}}}}}})} \right)}} = [/tex]

[tex]= \frac{{\ln {n^{\frac{1}{2}}} + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\ln {n^{\frac{1}{3}}} + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} = \frac{{\frac{1}{2}\ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3}\ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} =[/tex]

[tex]= \frac{{\frac{1}{2} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{2}}}}} + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{3}}}}} + \frac{1}{{{n^{\frac{1}{{12}}}}}})}}[/tex]

Now I'm stuck.

Is this the right way to do it?

Show the limit of

[tex]x_{n}=\frac{ln(1+\sqrt{n}+\sqrt[3]{n})}{ln(1+\sqrt[3]{n}+\sqrt[4]{n})}[/tex]

as n approaches infinity

Solution:

[tex]{x_n} = \frac{{\ln (1 + {n^{\frac{1}{2}}} + {n^{\frac{1}{3}}})}}{{\ln (1 + {n^{\frac{1}{3}}} + {n^{\frac{1}{4}}})}} = \frac{{\ln \left( {{n^{\frac{1}{2}}} \cdot (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})} \right)}}{{\ln \left( {{n^{\frac{1}{3}}} \cdot (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{\frac{1}{{12}}}}}})} \right)}} = [/tex]

[tex]= \frac{{\ln {n^{\frac{1}{2}}} + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\ln {n^{\frac{1}{3}}} + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} = \frac{{\frac{1}{2}\ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3}\ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} =[/tex]

[tex]= \frac{{\frac{1}{2} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{2}}}}} + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{3}}}}} + \frac{1}{{{n^{\frac{1}{{12}}}}}})}}[/tex]

Now I'm stuck.

Is this the right way to do it?