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Question Number 149346 by Study last updated on 04/Aug/21

$$li\underset{x\to \mathrm{\infty }}{m}{\left(\frac{3x+12}{6x-20}\right)}^{2x}=?$$

Commented by Study last updated on 04/Aug/21

$$helpeme$$

Answered by gsk2684 last updated on 04/Aug/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{3x+12}{6x-20}=\underset{x\to \mathrm{\infty }}{\lim }\frac{3+\frac{12}{x}}{6-\frac{20}{x}}=\frac{3+0}{6-0}=\frac{1}{2}<1$$$$then$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3x+12}{6x-20}\right)}^{2x}={\left(\frac{1}{2}\right)}^{\mathrm{\infty }}=0$$

Answered by puissant last updated on 04/Aug/21

$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\mathrm{e}}^{2\mathrm{xln}\left(\frac{3\mathrm{x}+12}{6\mathrm{x}-20}\right)}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\mathrm{e}}^{2\mathrm{xln}\left(\frac{3+\frac{12}{\mathrm{x}}}{6-\frac{20}{\mathrm{x}}}\right)}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\mathrm{e}}^{2\mathrm{xln}\left(\frac{1}{2}\right)}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{xln}2}$$$$=\{\begin{array}{l}+\mathrm{\infty }\mathrm{if}\mathrm{x}\to -\mathrm{\infty }\\ 0\mathrm{if}\mathrm{x}\to +\mathrm{\infty }\end{array}$$

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