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Question Number 120862 by bemath last updated on 03/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sin {}^2\left(\frac{2}{\mathrm{x}}\right)-\cos \left(\frac{1}{\mathrm{x}}\right)+1}{\sec \left(\frac{3}{\mathrm{x}}\right)\tan {}^2\left(\frac{3}{\mathrm{x}}\right)}?$$

Commented by liberty last updated on 03/Nov/20

$$\mathrm{take}\mathrm{X}=\frac{1}{\mathrm{x}}\mathrm{then}\underset{\mathrm{X}\to 0}{\lim }\frac{\sin {}^{2}2\mathrm{X}-\cos \mathrm{X}+1}{\sec 3\mathrm{X}.\tan {}^{2}3\mathrm{X}}=$$$$\underset{\mathrm{X}\to 0}{\lim }\cos {}^{3}3\mathrm{X}\times \underset{\mathrm{X}\to 0}{\lim }\frac{\sin {}^{2}2\mathrm{X}+2\sin {}^2\left(\mathrm{X}/2\right)}{\sin {}^{2}3\mathrm{X}}$$$$=1\times \frac{4+\frac{1}{2}}{9}=\frac{1}{2}$$

Answered by Dwaipayan Shikari last updated on 03/Nov/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{sin}^2\left(\frac{2}{x}\right)-cos\left(\frac{1}{x}\right)+1}{\frac{{sin}^2\left(\frac{3}{x}\right)}{{cos}^3\left(\frac{3}{x}\right)}}$$$$=\frac{{\left(\frac{2}{x}\right)}^2-1+\frac{1}{2x^2}+1}{{sin}^2\frac{3}{x}}=\frac{\frac{1}{x^2}(4+\frac{1}{2})}{{\left(\frac{3}{x}\right)}^2}=\frac{1}{2}$$

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