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Question-166560




Question Number 166560 by mathlove last updated on 22/Feb/22
Answered by qaz last updated on 22/Feb/22
lim_(x→0) ((ln(e^(sin x) +x^3 )−sin x)/(sin x−x(√(1−x^2 ))))  =lim_(x→0) ((ln((e^(sin x) +x^3 )/e^(sin x) ))/(x−(1/6)x^3 −x(1−(1/2)x^2 )+o(x^3 )))  =lim_(x→0) ((((e^(sin x) +x^3 )/e^(sin x) )−1)/((1/3)x^3 ))  =3
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{x}^{\mathrm{3}} \right)−\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}−\mathrm{x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\frac{\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{x}^{\mathrm{3}} }{\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} }}{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} −\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)+\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{x}^{\mathrm{3}} }{\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} }−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} } \\ $$$$=\mathrm{3} \\ $$

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