Question Number 179271 by cortano1 last updated on 27/Oct/22
![lim_(n→∞) [ ((n^3 +(n/3)))^(1/3) sin ((1/n)) ]^n^3 =?](https://www.tinkutara.com/question/Q179271.png)
$$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{\mathrm{n}^{\mathrm{3}} +\frac{\mathrm{n}}{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\:\right]^{\mathrm{n}^{\mathrm{3}} } =? \\ $$$$\:\: \\ $$
Answered by mr W last updated on 27/Oct/22
![=lim_(n→∞) [(1+(1/(3n^2 )))^(1/3) ×((sin (1/n))/(1/n))]^(1/n^3 ) =lim_(x→0) [(1+(x^2 /3))^(1/3) ×((sin x)/x)]^x^3 =(1×1)^0 =1](https://www.tinkutara.com/question/Q179272.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ×\frac{\mathrm{sin}\:\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}\right]^{\frac{\mathrm{1}}{{n}^{\mathrm{3}} }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ×\frac{\mathrm{sin}\:{x}}{{x}}\right]^{{x}^{\mathrm{3}} } \\ $$$$=\left(\mathrm{1}×\mathrm{1}\right)^{\mathrm{0}} \\ $$$$=\mathrm{1} \\ $$
Commented by Acem last updated on 28/Oct/22
$${Good}\:{work} \\ $$
Answered by cortano1 last updated on 28/Oct/22
![L = lim_(x→∞) [ ((n^3 +(n/3)))^(1/3) sin ((1/x))]^x^3 = lim_(t→0^+ ) [ (((1/t^3 )+(1/(3t))))^(1/3) sin t ]^(1/t^3 ) ln L = lim_(t→0^+ ) (((1/3)ln ((1/t^3 )+(1/(3t)))+ln (sin t))/t^3 ) ln L = lim_(t→0^+ ) ((ln ((1/t^3 )+(1/(3t)))+3ln (sin t))/(3t^3 )) ln L= lim_(t→0^+ ) ((ln (1+(t^2 /3))+3ln (((sin t)/t)))/(3t^3 )) ln L= lim_(t→0^+ ) (((t^2 /3)+O(t^4 )−3((t^2 /6)+O(t^4 )))/(3t^3 )) ln L = lim_(t→0^+ ) (−(1/(18t)) +O(t))=−∞ L=e^(−∞) = 0](https://www.tinkutara.com/question/Q179295.png)
$$\:\mathrm{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{\mathrm{n}^{\mathrm{3}} +\frac{\mathrm{n}}{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right]^{\mathrm{x}^{\mathrm{3}} } \\ $$$$\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3t}}}\:\mathrm{sin}\:\mathrm{t}\:\right]^{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }} \\ $$$$\:\mathrm{ln}\:\mathrm{L}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3t}}\right)+\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{t}\right)}{\mathrm{t}^{\mathrm{3}} }\: \\ $$$$\:\mathrm{ln}\:\mathrm{L}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3t}}\right)+\mathrm{3ln}\:\left(\mathrm{sin}\:\mathrm{t}\right)}{\mathrm{3t}^{\mathrm{3}} } \\ $$$$\:\mathrm{ln}\:\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{3}}\right)+\mathrm{3ln}\:\left(\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{t}}\right)}{\mathrm{3t}^{\mathrm{3}} } \\ $$$$\:\:\mathrm{ln}\:\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{O}\left(\mathrm{t}^{\mathrm{4}} \right)−\mathrm{3}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{6}}+\mathrm{O}\left(\mathrm{t}^{\mathrm{4}} \right)\right)}{\mathrm{3t}^{\mathrm{3}} } \\ $$$$\:\:\mathrm{ln}\:\mathrm{L}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(−\frac{\mathrm{1}}{\mathrm{18t}}\:+\mathrm{O}\left(\mathrm{t}\right)\right)=−\infty \\ $$$$\:\:\mathrm{L}=\mathrm{e}^{−\infty} \:=\:\mathrm{0}\: \\ $$
Commented by Acem last updated on 28/Oct/22
$$\:{Keep}\:{same}\:{variable} \\ $$