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Question Number 12513 by Joel577 last updated on 24/Apr/17

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

Answered by ajfour last updated on 24/Apr/17

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

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