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Question Number 188985 by normans last updated on 10/Mar/23

$$$$$${\mathbf{Lim}}_{\mathit{x}\to \sim }\frac{4{\mathit{x}}^3-2{\mathit{x}}^2-5\mathit{x}+4}{9{\mathit{x}}^3-4{\mathit{x}}^2+9}=??$$

Commented by MJS_new last updated on 10/Mar/23

$$\underset{x\to \pm \mathrm{\infty }}{\lim }\frac{{ax}^n+c_1{x}^{n-1}+...+c_n{x}^0}{{bx}^n+d_1{x}^{n-1}+...d_n{x}^0}=\frac{a}{b}$$

Answered by cortano12 last updated on 10/Mar/23

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{4\mathrm{x}}^3-{2\mathrm{x}}^2-5\mathrm{x}+4}{{9\mathrm{x}}^3-{4\mathrm{x}}^2+9}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^3(4-\frac{2}{\mathrm{x}}-\frac{5}{{\mathrm{x}}^2}+\frac{4}{{\mathrm{x}}^3})}{{\mathrm{x}}^3(9-\frac{4}{\mathrm{x}}+\frac{9}{{\mathrm{x}}^3})}$$$$=\frac{4}{9}$$

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