Question Number 12513 by Joel577 last updated on 24/Apr/17
![lim_(x→∞) x^2 [sec ((2/x)) − 1]](https://www.tinkutara.com/question/Q12513.png)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\left[\mathrm{sec}\:\left(\frac{\mathrm{2}}{{x}}\right)\:−\:\mathrm{1}\right] \\ $$
Answered by ajfour last updated on 24/Apr/17
![=lim_(x→∞) x^2 [(1/(cos (2/x)))−1] =lim_(x→∞) x^2 [((1−cos (2/x))/(cos (2/x)))] =lim_(x→∞) x^2 [((2sin^2 (1/x))/(cos (2/x)))] =lim_(x→∞) [(2)(((sin (1/x))/((1/x))))^2 .(1/(cos (2/x)))] = 2×1×(1/1) = 2 .](https://www.tinkutara.com/question/Q12514.png)
$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{2}/{x}\right)}−\mathrm{1}\right] \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left[\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{2}/{x}\right)}{\mathrm{cos}\:\left(\mathrm{2}/{x}\right)}\right] \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left[\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{1}/{x}\right)}{\mathrm{cos}\:\left(\mathrm{2}/{x}\right)}\right] \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(\mathrm{2}\right)\left(\frac{\mathrm{sin}\:\left(\mathrm{1}/{x}\right)}{\left(\mathrm{1}/{x}\right)}\right)^{\mathrm{2}} .\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{2}/{x}\right)}\right] \\ $$$$=\:\mathrm{2}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{1}}\:=\:\mathrm{2}\:. \\ $$