Đề sai nhé, x \to 1 chứ không phải x\to 2
Ta có
\dfrac{(\sqrt[3]{7x+1}-2) - ( \sqrt{5x-1}-2)}{x-1} =\dfrac{7(x-1)}{(x-1) (\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1} +4)}- \dfrac{5(x-1)}{(x-1) ( \sqrt{5x-1}+2)}
=\dfrac{7}{\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1} +4}- \dfrac{5}{ \sqrt{5x-1}+2}
Vậy \lim \limits_{x\to 1} \bigg ( \dfrac{7}{\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1} +4}- \dfrac{5}{ \sqrt{5x-1}+2} \bigg )=-\dfrac{4}{3}