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Question Number 116859 by Study last updated on 07/Oct/20

$$li\underset{x\to \mathrm{\infty }}{m}\frac{3x^2}{5^x}=?$$

Commented by JDamian last updated on 07/Oct/20

As exponential function increases more rapidly than parabola, you even can get this limit intuitively

Answered by bemath last updated on 07/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{6\mathrm{x}}{5^{\mathrm{x}}.\ln \left(5\right)}=\underset{x\to \mathrm{\infty }}{\lim }\frac{6}{5^{\mathrm{x}}.{\left(\ln \left(5\right)\right)}^2}=0$$

Commented by bemath last updated on 07/Oct/20

Answered by 1549442205PVT last updated on 07/Oct/20

$$li\underset{x\to \mathrm{\infty }}{m}\frac{3x^2}{5^x}=\mathrm{This}\mathrm{is}\mathrm{the}\mathrm{form}\frac{\mathrm{\infty }}{\mathrm{\infty }}.\mathrm{Hence},$$$$\mathrm{using}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule}\mathrm{we}\mathrm{have}$$$$\mathrm{I}=li\underset{x\to \mathrm{\infty }}{m}\frac{3x^2}{5^x}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{6\mathrm{x}}{5^{\mathrm{x}}\ln 5}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{6}{5^{\mathrm{x}}{\left(\ln 5\right)}^2}$$$$=0$$

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