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Question Number 168608 by Mastermind last updated on 14/Apr/22

Answered by infinityaction last updated on 14/Apr/22

Commented by SLVR last updated on 14/Apr/22

nice sir

$${nice}\:{sir} \\ $$$$ \\ $$

Answered by qaz last updated on 16/Apr/22

tan (e^(sin x) −sin x−1)=tan ((1/2)sin^2 x+(1/6)sin^3 x+...)=tan ((1/2)(x+...)^2 +(1/6)(x+...)^3 +...) =tan ((1/2)x^2 +(1/6)x^3 +...)=(1/2)x^2 +(1/6)x^3 +o(x^3 ) ln(2−cos x)=ln(1+(1/2)x^2 +...)=(1/2)x^2 +o(x^3 ) ⇒lim_(x→0) ((tan (e^(sin x) −sin x−1)−ln(2−cos x))/( ((1−x^3 ))^(1/5) −1))=lim_(x→0) (((1/6)x^3 +o(x^3 ))/(−(1/5)x^3 +o(x^3 )))=−(5/6)

$$\mathrm{tan}\:\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} −\mathrm{sin}\:\mathrm{x}−\mathrm{1}\right)=\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+...\right)=\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+...\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{x}+...\right)^{\mathrm{3}} +...\right) \\ $$$$=\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +...\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right) \\ $$$$\mathrm{ln}\left(\mathrm{2}−\mathrm{cos}\:\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +...\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} −\mathrm{sin}\:\mathrm{x}−\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{2}−\mathrm{cos}\:\mathrm{x}\right)}{\:\sqrt[{\mathrm{5}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)}{−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{x}^{\mathrm{3}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)}=−\frac{\mathrm{5}}{\mathrm{6}} \\ $$

Commented by SLVR last updated on 23/Apr/22

great sir

$${great}\:{sir} \\ $$

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