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Question Number 166111 by DAVONG last updated on 13/Feb/22

Answered by qaz last updated on 13/Feb/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{{\mathrm{x}}^2}\sin \mathrm{x}-\mathrm{x}(1+\frac{5}{6}{\mathrm{x}}^2)}{{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(1+{\mathrm{x}}^2+\frac{1}{2}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right))(\mathrm{x}-\frac{1}{6}{\mathrm{x}}^3+\frac{1}{120}{\mathrm{x}}^5+\mathrm{o}\left({\mathrm{x}}^5\right))-\mathrm{x}-\frac{5}{6}{\mathrm{x}}^3}{{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(\mathrm{x}+\frac{5}{6}{\mathrm{x}}^3+\frac{41}{120}{\mathrm{x}}^5+\mathrm{o}\left({\mathrm{x}}^5\right))-\mathrm{x}-\frac{5}{6}{\mathrm{x}}^3}{{\mathrm{x}}^5}$$$$=\frac{41}{120}$$

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