Question

∛5 x - 2 - 2 / √x - 1 - 1 com limite de X tendendo a 2?

Answer 1

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$\displaystyle\sf\lim_{x \to 2}\dfrac{\sqrt[3]{\sf 5x-2}-2}{\sqrt{x-1}-1}\\\displaystyle\sf\lim_{x \to 2}\dfrac{(\sqrt[3]{\sf 5x-2)}-2)}{(\sqrt{x-1}-1)}\cdot\dfrac{([\sqrt[3]{5x-2}~]^2+2\sqrt[3]{5x-2}+(-2)^2)}{([\sqrt[3]{\sf 5x-2}]^2+2\sqrt[3]{\sf5x-2}+(-2)^2)}\cdot\dfrac{(\sqrt{x-1}+1)}{(\sqrt{x-1}+1)}$

$\displaystyle\sf\lim_{x \to 2}\dfrac{(\sqrt[3]{\sf5x-2}^3-2^3)\cdot(\sqrt{x-1}+1)}{(\sqrt{x-1}^2-1^2)(\sqrt[3]{5x-2}^2+2\sqrt[3]{5x-2})+4)}$

$\displaystyle\sf\lim_{x \to 2}\dfrac{(\sf5x-2-8)(\sqrt{x-1}+1)}{(x-1-1)(\sqrt[3]{5x-2}^2+2\sqrt[3]{\sf5x-2}+4)}\\\displaystyle\sf\lim_{x \to 2}\dfrac{(5x-10)(\sqrt{x-1}+1)}{(x-2)(\sqrt[3]{\sf5x-2}^2+2\sqrt[3]{\sf5x-2}+4)}\\\displaystyle\sf5\lim_{x \to 2}\dfrac{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!2)(\sqrt{x-1}+1)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!2)(\sqrt[3]{5x-2}^2+2\sqrt[3]{5x-2}+4)}\\\sf =5\cdot\dfrac{\sqrt{2-1}+1}{(\sqrt[3]{\sf5\cdot 2-2})^2+2\sqrt[3]{\sf5\cdot2-2}+4}\\\sf =5\cdot\dfrac{2}{2^2+4+4}$

$\sf =5\cdot\dfrac{\backslash\!\!\!2^1}{\backslash\!\!\!8_4}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to 2}\dfrac{\sqrt[3]{\sf5x-2}-2}{\sqrt{x-1}-1}=\dfrac{5}{4}\checkmark}}}}}$