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Question Number 151538 by DELETED last updated on 21/Aug/21

Answered by DELETED last updated on 21/Aug/21

$$3).\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }[\left(\sin \frac{2}{\mathrm{t}}\right)-\frac{3}{\mathrm{t}}].\frac{\mathrm{t}}{6}=..?$$$$=\underset{\mathrm{t}\to 0}{\lim }[\left(\sin 2\mathrm{t}\right)-3\mathrm{t}].\frac{1}{6\mathrm{t}}$$$$=\underset{\mathrm{t}\to 0}{\lim }[\left(\frac{\sin 2\mathrm{t}}{6\mathrm{t}}\right)-\frac{1}{2}]$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\sin 2\mathrm{t}}{6\mathrm{t}}-\underset{\mathrm{t}\to 0}{\lim }\frac{1}{2}$$$$=\frac{2}{6}-\frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=-\frac{1}{6}//$$$$$$$$$$

Answered by DELETED last updated on 21/Aug/21

$$1).\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\mathrm{t}\sin \left(\frac{2}{\mathrm{t}}\right)=....?$$$$\mathrm{Jawab}:$$$$=\underset{\mathrm{y}\to 0}{\lim }\frac{1}{\mathrm{y}}\sin \left(2\mathrm{y}\right)$$$$=\underset{\mathrm{y}\to 0}{\lim }\frac{\sin \left(2\mathrm{y}\right)}{\mathrm{y}}=2//$$$$$$

Answered by DELETED last updated on 21/Aug/21

$$2).\underset{\theta \to \mathrm{\infty }}{\lim }{\theta }^2(1-\cos \frac{2}{\theta })=...?$$$$=\underset{\theta \to \mathrm{\infty }}{\lim }{\theta }^2[1-(1-2\sin {}^2\frac{1}{\theta }]$$$$=\underset{\theta \to \mathrm{\infty }}{\lim }{\theta }^2\left[2\sin {}^2\frac{1}{\theta }\right]$$$$=\underset{\theta \to 0}{\lim }\left(\frac{1}{{\theta }^2}\right)\left(2\sin {}^2\theta \right)$$$$=2\underset{\theta \to 0}{\lim }\left(\sin {}^2\theta \right)$$$$=2\underset{\theta \to 0}{\lim }\frac{\left(\sin {}^2\theta \right)}{{\theta }^2}=2\times 1=2//$$

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