Question

$\lim_{n \to 1/2} \sqrt[4]{2x-1}/ \sqrt{2x-1}$



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Answer 1

Bom día! 

$\lim_{x \to \dfrac{1}{2}}= \dfrac{ \sqrt[4]{2x-1} }{ \sqrt{2x-1}} \\ \\ \\ \lim_{x \to \dfrac{1}{2}}= \dfrac{ \sqrt[4]{2x-1} }{ \sqrt[2*2]{(2x-1)^2}} \\ \\ \\ \lim_{x \to \dfrac{1}{2}}= \dfrac{ \sqrt[4]{2x-1} }{ \sqrt[4]{(2x-1)^2}} \\ \\ \\ \lim_{x \to \dfrac{1}{2}}= \sqrt[4]{\dfrac{2x-1 }{(2x-1)^2}} \\ \\ \\ \lim_{x \to \dfrac{1}{2}}= \sqrt[4]{\dfrac{1 }{2x-1}} \\ \\ \\ \lim_{x \to \dfrac{1}{2}}= \sqrt[4]{\dfrac{1 }{2* \frac{1}{2}-1}}$


$\lim_{x \to \dfrac{1}{2}}= \sqrt[4]{\dfrac{1 }{0}} \\ \\ \\ \boxed{\lim_{x \to \dfrac{1}{2}}= \infty}$