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Question Number 132201 by SOMEDAVONG last updated on 12/Feb/21

$$\mathrm{A}=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\sin }^{50}\mathrm{x}+{5\sin }^{30}\mathrm{x}-6}{{\cos }^{20}\mathrm{x}+{\cos }^{40}2\mathrm{x}-1}$$

Answered by bemath last updated on 12/Feb/21

Answered by mathmax by abdo last updated on 13/Feb/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\sin }^{50}\mathrm{x}+{5\sin }^{30}\mathrm{x}-6}{{\cos }^{20}\mathrm{x}+{\cos }^{40}\mathrm{x}-1}\mathrm{changement}\mathrm{x}=\frac{\pi }{2}-\mathrm{t}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\frac{\pi }{2}-\mathrm{t})=\frac{{\cos }^{50}\mathrm{x}+{5\cos }^{30}\mathrm{x}-6}{{\sin }^{20}\mathrm{x}+{\cos }^{40}\left(2\mathrm{x}\right)-1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{(1-\frac{{\mathrm{x}}^2}{2})}^{50}+5{(1-\frac{{\mathrm{x}}^2}{2})}^{30}-6}{{\mathrm{x}}^{20}+{(1-{2\mathrm{x}}^2)}^{40}-1}\sim \frac{1-{25\mathrm{x}}^2+5(1-{15\mathrm{x}}^2)-6}{0+1-{80\mathrm{x}}^2-1}$$$$=\frac{-{25\mathrm{x}}^2-5.{15\mathrm{x}}^2}{-{80\mathrm{x}}^2}=\frac{25+5.15}{80}=\frac{100}{80}=\frac{10}{8}=\frac{5}{4}\Rightarrow {\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)=\frac{5}{4}$$$$$$

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