Question Number 173192 by mathlove last updated on 08/Jul/22
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)......\infty}}}=? \\ $$
Answered by Frix last updated on 08/Jul/22
![u=(√(t(√(t(√(t(√(...)))))))) ⇔ u=(√(tu)) ⇔ u=t lim_(n→+∞) [n^2 (1−cos (1/n))] = =lim_(k→0^+ ) [((1−cos k)/k^2 )] = =lim_(k→0^+ ) [((sin k)/(2k))] = =lim_(k→0^+ ) [((cos k)/2)] = =(1/2)](Q173200.png)
$${u}=\sqrt{{t}\sqrt{{t}\sqrt{{t}\sqrt{...}}}}\:\Leftrightarrow\:{u}=\sqrt{{tu}}\:\Leftrightarrow\:{u}={t} \\ $$$$ \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left[{n}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{{n}}\right)\right]\:= \\ $$$$=\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\frac{\mathrm{1}−\mathrm{cos}\:{k}}{{k}^{\mathrm{2}} }\right]\:= \\ $$$$=\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\frac{\mathrm{sin}\:{k}}{\mathrm{2}{k}}\right]\:= \\ $$$$=\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\frac{\mathrm{cos}\:{k}}{\mathrm{2}}\right]\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$