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Question Number 128422 by Khalmohmmad last updated on 07/Jan/21

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n^a}{b^n}a>0,b>1$$

Answered by mathmax by abdo last updated on 07/Jan/21

$$\mathrm{let}{\mathrm{u}}_{\mathrm{n}}=\frac{{\mathrm{n}}^{\mathrm{a}}}{{\mathrm{b}}^{\mathrm{n}}}\Rightarrow \ln \left({\mathrm{u}}_{\mathrm{n}}\right)=\mathrm{aln}\left(\mathrm{n}\right)-\mathrm{nlnb}=\mathrm{n}(\frac{\mathrm{aln}\left(\mathrm{n}\right)}{\mathrm{n}}-\ln \left(\mathrm{b}\right))$$ $$\mathrm{we}\mathrm{have}\mathrm{b}>1\Rightarrow \mathrm{lnb}>0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{u}}_{\mathrm{n}}\right)=-\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$

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