Question Number 55334 by Mikael_Marshall last updated on 21/Feb/19
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{8}^{{n}} }{\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{3}^{{n}+\mathrm{2}} } \\ $$
Answered by mr W last updated on 21/Feb/19
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} +\mathrm{9}\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{0}+\mathrm{9}×\mathrm{0}} \\ $$$$=\infty \\ $$