Question Number 95133 by john santu last updated on 23/May/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{x}^{\mathrm{2}} .\mathrm{e}^{\:−\mathrm{2x}} \right)\:=\:? \\ $$
Answered by bobhans last updated on 23/May/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{2x}} }\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{2e}^{\mathrm{2x}} }\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{x}}{\mathrm{e}^{\mathrm{2x}} }\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2e}^{\mathrm{2x}} }\:=\:\mathrm{0} \\ $$
Commented by john santu last updated on 23/May/20
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