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Question Number 98833 by john santu last updated on 16/Jun/20

Commented by bobhans last updated on 16/Jun/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(3^{\mathrm{n}}{(1+\left(\frac{1}{6^{\mathrm{n}}}\right))}^{\frac{1}{\mathrm{n}}}\right)=3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{(1+\frac{1}{6^{\mathrm{n}}})}^{\frac{1}{\mathrm{n}}}$$$$=3\times 1=3$$

Answered by mathmax by abdo last updated on 16/Jun/20

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={(3^{\mathrm{n}}+\frac{1}{2^{\mathrm{n}}})}^{\frac{1}{\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=3{(1+\frac{1}{6^{\mathrm{n}}})}^{\mathrm{n}}=3{\mathrm{e}}^{\mathrm{nln}(1+\frac{1}{6^{\mathrm{n}}})}\mathrm{but}$$$$\ln (1+\frac{1}{6^{\mathrm{n}}})\sim \frac{1}{6^{\mathrm{n}}}\Rightarrow \mathrm{n}\mathrm{on}(1+\frac{1}{6^{\mathrm{n}}})\sim \frac{\mathrm{n}}{6^{\mathrm{n}}}\to 0(\mathrm{n}\to +\mathrm{\infty })\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=3$$

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