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Question Number 114020 by bemath last updated on 16/Sep/20

$$\underset{d\to \mathrm{\infty }}{\lim }\sqrt{6d}\cos \left(\frac{2}{\sqrt{d}}\right)\sin \left(\frac{5}{\sqrt{d}}\right)?$$$$$$

Answered by Olaf last updated on 17/Sep/20

$$\sqrt{6d}\cos \left(\frac{2}{\sqrt{d}}\right)\sin \left(\frac{5}{\sqrt{d}}\right)\underset{\mathrm{\infty }}{\sim }\sqrt{6d}(1-\frac{2}{d})\left(\frac{5}{\sqrt{d}}\right)$$$$\underset{d\to \mathrm{\infty }}{\to }5\sqrt{6}$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{6\mathrm{x}}\cos \left(\frac{2}{\sqrt{\mathrm{x}}}\right).\sin \left(\frac{5}{\sqrt{\mathrm{x}}}\right)(\mathrm{d}\to +\mathrm{\infty })\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \sqrt{6}\sqrt{\mathrm{x}}(1-\frac{4}{\mathrm{x}}).\frac{5}{\sqrt{\mathrm{x}}}=5\sqrt{6}\{1-\frac{4}{\mathrm{x}}\}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=5\sqrt{6}$$

Answered by bobhans last updated on 17/Sep/20

$$letting\frac{1}{\sqrt{d}}=q\wedge q\to 0$$$$\underset{q\to 0}{\lim }\frac{\sqrt{6}}{q}.\cos 2q.\sin 5q=\underset{q\to 0}{\lim }\sqrt{6}\cos 2q.\underset{q\to 0}{\lim }\frac{\sin 5q}{q}$$$$=\sqrt{6}\times 5=5\sqrt{6}$$

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