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Question Number 184341 by Davidtim last updated on 05/Jan/23

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

Commented by JDamian last updated on 05/Jan/23

zero, as the denominator tends to infinity faster than the numerator does.

Answered by SEKRET last updated on 05/Jan/23

$$\underset{\mathbf{x}\to \mathrm{\infty }}{\mathbf{lim}}\frac{{\mathbf{x}}^2}{3^{\mathbf{x}}}=\underset{\mathbf{x}\to \mathrm{\infty }}{\mathbf{lim}}\frac{2\mathbf{x}}{3^{\mathbf{x}}\mathbf{ln}\left(3\right)}=\underset{\mathbf{x}\to \mathrm{\infty }}{\mathbf{lim}}\frac{2}{3^{\mathbf{x}}{\mathbf{ln}}^2\left(3\right)}=0$$

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