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Question Number 159720 by tounghoungko last updated on 20/Nov/21

L = lim_(x→(π/3)) ((3−4sin^2 x)/(sin 2x−sin x)) ? Q = lim_(x→0) [(1/x^2 ) ((2/(cos^2 x)) +cos x−3)] ?

$$\:\:\:\:\:\:\:\:{L}\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:{x}}\:? \\ $$$$\:\:\:\:\:\:{Q}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\left(\frac{\mathrm{2}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\mathrm{cos}\:{x}−\mathrm{3}\right)\right]\:?\: \\ $$

Commented by cortano last updated on 20/Nov/21

L = lim_(x→(π/3)) ((3−4sin^2 x)/(sin 2x−sin x)) L= lim_(x→(π/3)) ((3−4(1−cos^2 x))/(sin x(2cos x−1))) L= lim_(x→(π/3)) ((4cos^2 x−1)/(sin x(2cos x−1))) L= lim_(x→(π/3)) ((2cos x+1)/(sin x)) = (2/((√3)/2)) = (4/( (√3)))

$$\:\:\:\:\:\:{L}\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:{x}} \\ $$$$\:\:\:\:\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{sin}\:{x}\left(\mathrm{2cos}\:{x}−\mathrm{1}\right)} \\ $$$$\:\:\:\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{sin}\:{x}\left(\mathrm{2cos}\:{x}−\mathrm{1}\right)} \\ $$$$\:\:\:\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}+\mathrm{1}}{\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{2}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Answered by blackmamba last updated on 20/Nov/21

determinant (((Q=lim_(x→0) (((cos^3 x−3cos^2 x+2)/(x^2 cos^2 x))))),((Q= lim_(x→0) (((cos x−1)/x^2 ))(((cos^2 x−2cos x−2)/(cos^2 x))))),((Q=−3.(−2)((1/4))= 1.5 )))

$$\:\begin{array}{|c|c|c|}{{Q}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:^{\mathrm{2}} {x}+\mathrm{2}}{{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} {x}}\right)}\\{{Q}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:{x}−\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left(\frac{\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2cos}\:{x}−\mathrm{2}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\right)}\\{{Q}=−\mathrm{3}.\left(−\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\:\mathrm{1}.\mathrm{5}\:}\\\hline\end{array} \\ $$

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