Question Number 139059 by sahnaz last updated on 21/Apr/21
Answered by mr W last updated on 21/Apr/21
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}×\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\mathrm{9}×\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{64}}{\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{2}×\mathrm{0}+\mathrm{9}×\mathrm{0}+\mathrm{64}}{\mathrm{0}+\mathrm{0}+\mathrm{1}} \\ $$$$=\frac{\mathrm{64}}{\mathrm{1}} \\ $$$$=\mathrm{64} \\ $$