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Question Number 193976 by Risandu last updated on 25/Jun/23

Answered by Subhi last updated on 25/Jun/23

$$$$$${lim}_{x\to 1}\frac{{({}^3\sqrt{x}-1)}^2}{{(x-1)}^2}$$$${lim}_{x\to 1}\frac{{({}^3\sqrt{x}-1)}^2}{{({}^3\sqrt{x}-1)}^2{({}^3\sqrt{x^2}{+}^3\sqrt{x}+1)}^2}\Rrightarrow {lim}_{x\to 1}\frac{1}{{({}^3\sqrt{x^2}{+}^3\sqrt{x}+1)}^2}=\frac{1}{9}$$$$or$$$${lim}_{x\to 1}\frac{\frac{2}{3}{x}^{\frac{-1}{3}}-\frac{2}{3}{x}^{-\frac{2}{3}}}{2(x-1)}\left\{L{Hopital}^{\prime }slaw\right\}$$$${lim}_{x\to 1}\frac{\frac{-2}{9}{x}^{\frac{-4}{3}}+\frac{4}{9}{x}^{\frac{-5}{3}}}{2}=\frac{1}{9}$$

Answered by cortano12 last updated on 25/Jun/23

$$\mathrm{L}={\left[\underset{x\to 1}{\lim }\frac{\sqrt[3]{\mathrm{x}}-1}{\mathrm{x}-1}\right]}^2$$$$\mathrm{L}={\left[\underset{x\to 1}{\lim }\frac{1}{{\left(\sqrt[3]{\mathrm{x}}\right)}^2+\sqrt[3]{\mathrm{x}}+1}\right]}^2$$$$\mathrm{L}=\overline{)\begin{array}{c}\frac{1}{9}\end{array}}$$$$$$

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