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Question Number 95133 by john santu last updated on 23/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }({\mathrm{x}}^2.{\mathrm{e}}^{-2\mathrm{x}})=?$$

Answered by bobhans last updated on 23/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{2\mathrm{x}}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{2\mathrm{x}}{{2\mathrm{e}}^{2\mathrm{x}}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}}{{\mathrm{e}}^{2\mathrm{x}}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{2\mathrm{e}}^{2\mathrm{x}}}=0$$

Commented by john santu last updated on 23/May/20

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