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Question Number 33312 by abdo imad last updated on 14/Apr/18
calculate lim_(x→0)    ((e^(−3x^2 )  −1)/x^2 ) .
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \:−\mathrm{1}}{{x}^{\mathrm{2}} }\:. \\ $$
Answered by Joel578 last updated on 14/Apr/18
lim_(x→0)  ((e^(−3x^2 )  − 1)/x^2 )  (L′Hopital)   = lim_(x→0)  ((−3)/e^(3x^2 ) ) = −3
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \:−\:\mathrm{1}}{{x}^{\mathrm{2}} }\:\:\left({L}'{Hopital}\right) \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{3}}{{e}^{\mathrm{3}{x}^{\mathrm{2}} } }\:=\:−\mathrm{3} \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
we have e^u   =1  +u  +o(u^2 )(u→0) ⇒  e^(−3x^2 )  =1−3x^2  +o(x^4 ) ⇒ e^(−3x^2 ) −1 = −3x^2  +o(x^4 )  ⇒ ((e^(−3x^2 ) −1)/x^2 ) =−3 +o(x^2 ) ⇒lim_(x→0)  ((e^(−3x^2 ) −1)/x^2 ) =−3 .
$${we}\:{have}\:{e}^{{u}} \:\:=\mathrm{1}\:\:+{u}\:\:+{o}\left({u}^{\mathrm{2}} \right)\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${e}^{−\mathrm{3}{x}^{\mathrm{2}} } \:=\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow\:{e}^{−\mathrm{3}{x}^{\mathrm{2}} } −\mathrm{1}\:=\:−\mathrm{3}{x}^{\mathrm{2}} \:+{o}\left({x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:\frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } −\mathrm{1}}{{x}^{\mathrm{2}} }\:=−\mathrm{3}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } −\mathrm{1}}{{x}^{\mathrm{2}} }\:=−\mathrm{3}\:. \\ $$

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