Question Number 200304 by Calculusboy last updated on 16/Nov/23
Answered by Sutrisno last updated on 17/Nov/23
$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\mathrm{0}.\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}} \\ $$
Commented by Calculusboy last updated on 28/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by MathematicalUser2357 last updated on 05/Jan/24
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}0}.\mathrm{2}^{\mathrm{log}_{\sqrt{\mathrm{5}}} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}0}.\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$