Question Number 219801 by SdC355 last updated on 02/May/25
$$\underset{{h}\rightarrow\infty} {\mathrm{lim}}\:{h}^{\nu} {J}_{\nu} \left({h}\right)=?? \\ $$
Answered by MrGaster last updated on 02/May/25
$$\mathrm{lim}_{{h}\rightarrow\infty} \:{h}^{\nu} {J}_{\nu} \left({h}\right)=\sqrt{\frac{\mathrm{2}}{\pi}}\mathrm{lim}_{{h}\rightarrow\infty} {h}^{\nu−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{cos}\left({h}−\frac{\nu\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\begin{cases}{\mathrm{if}\:\nu<\frac{\mathrm{1}}{\mathrm{2}},{h}^{\nu−\frac{\mathrm{1}}{\mathrm{2}}} \rightarrow\mathrm{0}\:}\\{\mathrm{if}\:\nu\geq\frac{\mathrm{1}}{\mathrm{2}}\:\nexists\underline{\mathrm{lim}}}\end{cases} \\ $$