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Question Number 112917 by abdullahquwatan last updated on 10/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }[{\csc }^2\left(\frac{2}{\mathrm{x}}\right)-\frac{1}{4}{\mathrm{x}}^2]$$

Commented by bobhans last updated on 10/Sep/20

$$\underset{\mathrm{w}\to 0}{\lim }\left(\frac{2\mathrm{w}+\sin \left(2\mathrm{w}\right)}{4\mathrm{w}}\right)\left(\frac{2\mathrm{w}-\sin \left(2\mathrm{w}\right)}{\mathrm{w}.{\sin }^2\left(2\mathrm{w}\right)}\right)$$$$1\times \underset{\mathrm{w}\to 0}{\lim }\left(\frac{2\mathrm{w}-\sin 2\mathrm{w}}{{4\mathrm{w}}^3}\right)=$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{2-2\cos 2\mathrm{w}}{{12\mathrm{w}}^2}=\underset{\mathrm{w}\to 0}{\lim }\left(\frac{1-\cos 2\mathrm{w}}{{6\mathrm{w}}^2}\right)$$$$=\underset{\mathrm{w}\to 0}{\lim }\left(\frac{2\sin {}^2\mathrm{w}}{{6\mathrm{w}}^2}\right)=\frac{1}{3}$$

Commented by abdullahquwatan last updated on 10/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by abdullahquwatan last updated on 10/Sep/20

$$\mathrm{why}{4\mathrm{w}}^3??$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\sin }^2\left(\frac{2}{\mathrm{x}}\right)}-\frac{{\mathrm{x}}^2}{4}\mathrm{changement}\frac{2}{\mathrm{x}}=\mathrm{t}\mathrm{give}\frac{\mathrm{x}}{2}=\frac{1}{\mathrm{t}}\Rightarrow \mathrm{x}=\frac{2}{\mathrm{t}}$$$$(\mathrm{x}\to \mathrm{\infty }\iff \mathrm{t}\to 0)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\sin }^2\mathrm{t}}-\frac{1}{4}\left(\frac{4}{{\mathrm{t}}^2}\right)=\frac{1}{{\sin }^2\mathrm{t}}-\frac{1}{{\mathrm{t}}^2}=\mathrm{g}\left(\mathrm{t}\right)$$$$=\frac{{\mathrm{t}}^2-{\sin }^2\mathrm{t}}{{\mathrm{t}}^2{\sin }^2\mathrm{t}}\mathrm{we}\mathrm{have}\mathrm{sint}\sim \mathrm{t}-\frac{{\mathrm{t}}^3}{6}\Rightarrow {\sin }^2\mathrm{t}\sim {(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})}^2={\mathrm{t}}^2-\frac{1}{3}{\mathrm{t}}^4+\frac{{\mathrm{t}}^6}{36}$$$$\sim {\mathrm{t}}^2-\frac{1}{3}{\mathrm{t}}^4\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim \frac{{\mathrm{t}}^2-{\mathrm{t}}^2+\frac{1}{3}{\mathrm{t}}^4}{{\mathrm{t}}^2.{\mathrm{t}}^2}=\frac{1}{3}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=\frac{1}{3}\Rightarrow$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{3}$$

Commented by abdullahquwatan last updated on 10/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 11/Sep/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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