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Question Number 9768 by ridwan balatif last updated on 01/Jan/17

lim_(x→0) ((sin3xcos3x−sin3x)/(2x^2 tan2xcos(1/2)x))=...?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin3}{x}\mathrm{cos3}{x}−\mathrm{sin3}{x}}{\mathrm{2}{x}^{\mathrm{2}} \mathrm{tan2}{x}\mathrm{cos}\frac{\mathrm{1}}{\mathrm{2}}{x}}=...? \\ $$

Answered by sandy_suhendra last updated on 01/Jan/17

=lim_(x→0) ((sin3x (cos3x−1))/(2x^2 tan2x cos(1/2)x)) =lim_(x→0) ((sin3x (−2sin^2 1(1/2)x))/(2x^2 tan2x cos(1/2)x)) =lim_(x→0) ((sin3x)/(tan 2x))×((−2sin^2 1(1/2)x)/(2x^2 ))×(1/(cos 0)) =− (3/2)×(1(1/2))^2 × 1 = − ((27)/8)

$$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{sin3x}\:\left(\mathrm{cos3x}−\mathrm{1}\right)}{\mathrm{2x}^{\mathrm{2}} \:\mathrm{tan2x}\:\mathrm{cos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}} \\ $$$$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{sin3x}\:\left(−\mathrm{2sin}^{\mathrm{2}} \:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{2x}^{\mathrm{2}} \:\mathrm{tan2x}\:\mathrm{cos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}} \\ $$$$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{sin3x}}{\mathrm{tan}\:\mathrm{2x}}×\frac{−\mathrm{2sin}^{\mathrm{2}} \:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{2x}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{0}} \\ $$$$=−\:\frac{\mathrm{3}}{\mathrm{2}}×\left(\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\:\mathrm{1} \\ $$$$=\:−\:\frac{\mathrm{27}}{\mathrm{8}} \\ $$

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