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lim-3-2-8-16-2-2-9-4-




Question Number 204313 by serenity last updated on 12/Feb/24
lim((3×^2 −8×−16)/(2×^2 9×+4))
$${lim}\frac{\mathrm{3}×^{\mathrm{2}} −\mathrm{8}×−\mathrm{16}}{\mathrm{2}×^{\mathrm{2}} \mathrm{9}×+\mathrm{4}} \\ $$$$ \\ $$
Answered by Faetmaaa last updated on 27/Feb/24
lim_(x→∞)  ((3x^2 −8x−16)/(2x^2 +9x+4))  = lim_(x→∞)  ((3−(8/x)−((16)/x^2 ))/(2+(9/x)+(4/x^2 )))  = ((3−0−0)/(2+0+0))  = (3/2)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{16}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{4}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\frac{\mathrm{8}}{{x}}−\frac{\mathrm{16}}{{x}^{\mathrm{2}} }}{\mathrm{2}+\frac{\mathrm{9}}{{x}}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{3}−\mathrm{0}−\mathrm{0}}{\mathrm{2}+\mathrm{0}+\mathrm{0}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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