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Question Number 193596 by Mingma last updated on 17/Jun/23

Answered by cortano12 last updated on 17/Jun/23

$$\left(1\right)\gamma =1$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-2\alpha \mathrm{x}-\beta }{2\mathrm{x}}=\frac{3}{2}$$$$\beta =1$$$$\left(3\right)\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-2\alpha }{2}=\frac{3}{2}$$$$\alpha =-1$$$$\therefore \underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2-\mathrm{x}-1}{{\mathrm{x}}^2}=\frac{3}{2}$$

Commented by Mingma last updated on 17/Jun/23

Perfect ��

Answered by mnjuly1970 last updated on 17/Jun/23

$$lim\frac{1+x+\frac{x^2}{2}+o\left(x^3\right)-\alpha x^2-\beta x-\gamma }{x^2}=\frac{3}{2}$$$$\gamma =1,\beta =1$$$$\frac{1}{2}-\alpha =\frac{3}{2}\Rightarrow \alpha =-1$$

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