Question Number 149377 by fotosy2k last updated on 05/Aug/21 | ||
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Answered by john_santu last updated on 05/Aug/21 | ||
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$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}.\mathrm{3}^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}} \left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} \right)}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} }=\mathrm{3} \\ $$ | ||
Commented by fotosy2k last updated on 05/Aug/21 | ||
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$${thank}\:{you} \\ $$$$ \\ $$ | ||