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Question Number 33312 by abdo imad last updated on 14/Apr/18

$$calculate{lim}_{x\to 0}\frac{e^{-3x^2}-1}{x^2}.$$

Answered by Joel578 last updated on 14/Apr/18

$$\underset{x\to 0}{\lim }\frac{e^{-3x^2}-1}{x^2}\left(L^{\prime }Hopital\right)$$$$=\underset{x\to 0}{\lim }\frac{-3}{e^{3x^2}}=-3$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$wehave{e}^u=1+u+o\left(u^2\right)(u\to 0)\Rightarrow$$$$e^{-3x^2}=1-3x^2+o\left(x^4\right)\Rightarrow e^{-3x^2}-1=-3x^2+o\left(x^4\right)$$$$\Rightarrow \frac{e^{-3x^2}-1}{x^2}=-3+o\left(x^2\right)\Rightarrow {lim}_{x\to 0}\frac{e^{-3x^2}-1}{x^2}=-3.$$

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