Question Number 162790 by LEKOUMA last updated on 01/Jan/22
$${Calculate}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{2}^{{n}} +\mathrm{3}^{{n}} \right)^{\frac{\mathrm{1}}{{n}}} \\ $$
Answered by mindispower last updated on 01/Jan/22
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{\frac{{ln}\left(\mathrm{3}^{{n}} \right)+{ln}\left(\frac{\mathrm{1}}{\mathrm{3}^{{n}} }+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} +\mathrm{1}\right)}{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{{ln}\left(\mathrm{3}\right)} .{e}^{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \right)} =\mathrm{3} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\rightarrow\mathrm{0},\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \rightarrow\mathrm{0} \\ $$