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Question Number 146559 by liberty last updated on 14/Jul/21

$$\underset{x\to -\mathrm{\infty }}{\lim }(-2x+1+\sqrt[3]{8x^3+2x^2-10})=?$$$$\underset{x\to \mathrm{\infty }}{\lim }(\sqrt[3]{125x^3-5x^2}-4-5x)=?$$

Answered by iloveisrael last updated on 14/Jul/21

$$\overline{)\begin{array}{c}\left(1\right)\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1+2(\mathrm{x}+\frac{2}{3.8}))=1+\frac{1}{6}=\frac{7}{6}\\ \left(2\right)\underset{x\to \mathrm{\infty }}{\lim 5}(\mathrm{x}-\frac{5}{3.125})-5\mathrm{x}-4=-\frac{1}{15}-4=-\frac{61}{15}\end{array}}$$

Answered by liberty last updated on 14/Jul/21

$$\left(1\right)\underset{x\to -\mathrm{\infty }}{\lim }x(-2+\frac{1}{x})+x\sqrt[3]{8+\frac{2}{x}-\frac{10}{x^3}}$$$$=\underset{t\to 0}{\lim }\frac{-2+t+\sqrt[3]{8+2t-10t^3}}{t}$$$$=\underset{t\to 0}{\lim }\frac{1+\frac{2-30t^2}{3(\sqrt[3]{{(8+2t-10t^3)}^2}}}{1}$$$$=1+\frac{2}{3.4}=\frac{7}{6}$$

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