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Question Number 83329 by john santu last updated on 01/Mar/20

$$\underset{x\to 0}{\lim }\frac{({\mathrm{e}}^{-2\mathrm{x}}-(1+\mathrm{ax}))}{{\mathrm{x}}^2(1+\mathrm{bx})}=?$$

Commented by jagoll last updated on 01/Mar/20

$$\underset{x\to 0}{\lim }\frac{1}{1+\mathrm{bx}}\times \underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{-2\mathrm{x}}-1-\mathrm{ax}}{{\mathrm{x}}^2}=?$$$$1\times \underset{x\to 0}{\lim }\frac{-{2\mathrm{e}}^{-2\mathrm{x}}-\mathrm{a}}{2\mathrm{x}}$$$$\mathrm{if}\mathrm{a}\ne -2\Rightarrow \underset{x\to 0}{\lim }\frac{-{2\mathrm{e}}^{-2\mathrm{x}}-\mathrm{a}}{2\mathrm{x}}=\mathrm{\infty }$$$$\mathrm{if}\mathrm{a}=-2\Rightarrow \underset{x\to 0}{\lim }\frac{-{2\mathrm{e}}^{-2\mathrm{x}}+2}{2\mathrm{x}}=$$$$\underset{x\to 0}{\lim }\frac{{4\mathrm{e}}^{-2\mathrm{x}}}{2}=2$$

Answered by jagoll last updated on 01/Mar/20

$$\underset{x\to 0}{\lim }\frac{-{2\mathrm{e}}^{-2\mathrm{x}}-\mathrm{a}}{2\mathrm{x}(1+\mathrm{bx})+{\mathrm{bx}}^2},\mathrm{a}\mathrm{must}\mathrm{be}=-2$$$$\underset{x\to 0}{\lim }\frac{-{2\mathrm{e}}^{-2\mathrm{x}}+2}{2\mathrm{x}+{3\mathrm{bx}}^2}=\underset{x\to 0}{\lim }\frac{{4\mathrm{e}}^{-2\mathrm{x}}}{2+6\mathrm{bx}}=2$$

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