# 1-lim-n-3-n-2-n-6-n-and-lim-n-1-2-2-32-4-2-n-2-n-3-help-me-

Question Number 131202 by abdurehime last updated on 02/Feb/21
$$\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \left(\frac{\mathrm{3}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} }{\mathrm{6}^{\mathrm{n}} }\right)\:\mathrm{and} \\$$$$\:\:\:\overset{\:\mathrm{lim}\:\underset{\mathrm{n}\rightarrow\infty} {\:}\left(\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{32}+\mathrm{4}^{\mathrm{2}} +…….+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right)} {\:} \\$$$$\mathrm{help}\:\mathrm{me} \\$$
Commented by liberty last updated on 02/Feb/21
$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +…+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right)\: \\$$
Answered by liberty last updated on 02/Feb/21
$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6}^{\mathrm{n}} \left(\left(\frac{\mathrm{3}}{\mathrm{6}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{6}}\right)^{\mathrm{n}} \right)}{\mathrm{6}^{\mathrm{n}} }\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{n}} \right] \\$$$$\:=\:\mathrm{0}\:+\:\mathrm{0}\:=\:\mathrm{0} \\$$
Commented by EDWIN88 last updated on 03/Feb/21
$${it}\:{is}\:{right}\:{answer}\:! \\$$
Answered by Chhing last updated on 02/Feb/21
$$\\$$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{n}^{\mathrm{3}} }\right) \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)}{\mathrm{n}^{\mathrm{3}} }\right) \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2} \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}} \\$$
Answered by Ar Brandon last updated on 02/Feb/21
$$\mathrm{1}.\mathscr{L}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} }{\mathrm{6}^{\mathrm{n}} }\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\left(\frac{\mathrm{3}}{\mathrm{6}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{6}}\right)^{\mathrm{n}} \right\}=\mathrm{0} \\$$$$\mathrm{2}.\mathscr{T}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{n}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }\right\}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} } \\$$$$\:\:\:\:\:\:\:\:=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right\}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{3}} \\$$$$\\$$