Question Number 57957 by rahul 19 last updated on 15/Apr/19
Commented by Smail last updated on 15/Apr/19
$${L}_{{n}} =\frac{\mathrm{2}^{{n}} +\left(−\mathrm{2}\right)^{{n}} }{\mathrm{3}^{{n}} }=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$${As}\:{n}\rightarrow\infty,\:\:{L}_{{n}} \rightarrow\mathrm{0}\:\:{because}\:\:\:\frac{\mathrm{2}}{\mathrm{3}}<\mathrm{1} \\ $$$${However}\:\: \\ $$$${l}_{{n}} =\frac{\mathrm{2}^{{n}} +\left(−\mathrm{2}\right)^{{n}} }{\mathrm{2}^{{n}} }=\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \\ $$$${If}\:{n}\:{is}\:{even},\:{then}\:\:\underset{{n}\rightarrow\infty} {{lim}l}_{{n}} =\mathrm{2}\: \\ $$$${If}\:{not}\:\:\underset{{n}\rightarrow\infty} {{lim}l}_{{n}} =\mathrm{0} \\ $$$${Which}\:\:{means}\:\:\underset{{n}\rightarrow\infty} {{lim}l}_{{n}} {does}\:{not}\:{exist}. \\ $$
Commented by rahul 19 last updated on 16/Apr/19
$${Thank}\:{U}\:{Sir}. \\ $$