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Question Number 93695 by i jagooll last updated on 14/May/20

Commented by mathmax by abdo last updated on 14/May/20

$$f\left(x\right)=\frac{6x{-}^3\sqrt{27x^3+2x-1}}{({}^3\sqrt{8x^3-x}+x)}\Rightarrow f\left(x\right)=\frac{6x-3x{(1+\frac{2}{27x^2}-\frac{1}{27x^3})}^{\frac{1}{3}}}{2x{(1-\frac{1}{8x^2})}^{\frac{1}{3}}+x}$$$$\Rightarrow f\left(x\right)\sim \frac{6x-3x(1+\frac{1}{3}(\frac{2}{27x^2}-\frac{1}{27x^3}))}{2x(1-\frac{1}{24x^2})+x}\sim \frac{3x}{3x}=1\Rightarrow {lim}_{x\to -\mathrm{\infty }}f\left(x\right)=1$$

Answered by 1549442205 last updated on 14/May/20

$$\mathbf{li}\underset{\mathbf{x}\to -\mathrm{\infty }}{\mathbf{m}}\frac{6\mathbf{x}-\sqrt[3]{27{\mathbf{x}}^3+2\mathbf{x}-1}}{\sqrt[3]{8{\mathbf{x}}^3-\mathbf{x}}+\mathrm{x}}=\mathbf{li}\underset{\mathbf{x}\to -\mathrm{\infty }}{\mathbf{m}}\frac{6-\sqrt[3]{27+\frac{2}{{\mathrm{x}}^2}-\frac{1}{{\mathrm{x}}^3}}}{\sqrt[3]{8-\frac{1}{{\mathrm{x}}^2}}+1}$$$$=\frac{6-\sqrt[3]{27}}{\sqrt[3]{8}+1}=\frac{6-3}{2+1}=1$$

Commented by john santu last updated on 14/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\ne \underset{x\to -\mathrm{\infty }}{\lim }\mathrm{sir}$$

Commented by john santu last updated on 14/May/20

$$\mathrm{your}\mathrm{answer}\mathrm{is}\mathrm{wrong}$$

Commented by 1549442205 last updated on 14/May/20

$$\mathrm{excuse}\mathrm{me},\mathrm{I}\mathrm{wrote}\mathrm{fault}$$

Answered by john santu last updated on 14/May/20

$$\underset{x\to -\mathrm{\infty }}{\lim }\left(\frac{6\mathit{x}-\mathit{x}\sqrt[3]{27+\frac{2}{{\mathit{x}}^2}-\frac{1}{{\mathit{x}}^3}}}{\mathit{x}\sqrt[3]{8-\frac{1}{{\mathit{x}}^2}}+\mathit{x}}\right)=$$$$\underset{x\to -\mathrm{\infty }}{\lim }\left(\frac{-6+3}{-2-1}\right)=\frac{-3}{-3}=1$$

Commented by i jagooll last updated on 14/May/20

cooll man ������

Commented by Ar Brandon last updated on 14/May/20

My man��

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