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Question Number 190765 by mustafazaheen last updated on 10/Apr/23

Answered by qaz last updated on 11/Apr/23

$$\underset{x\to 0}{lim}\frac{-e^2+{(1+2x)}^{\frac{1}{x}}}{\sin x}=\underset{x\to 0}{lim}\frac{-e^2+e^{\frac{1}{x}ln(1+2x)}}{x}$$$$=\underset{x\to 0}{lim}\frac{-e^2+e^{2-2x+o\left(x\right)}}{x}=\underset{x\to 0}{lim}\frac{-e^2+e^2(1-2x+o\left(x\right))}{x}=-2e^2$$

Commented by mustafazaheen last updated on 12/Apr/23

$$\mathrm{Thanks}\mathrm{Sir}$$$$\mathrm{o}\left(\mathrm{x}\right)=?$$

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