Tính \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}\).
\(\dfrac{3}{10}\).\(\dfrac{5}{10}\).\(\dfrac{7}{10}\).\(\dfrac{9}{10}\).Hướng dẫn giải:Có \(\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}=\frac{\sqrt[4]{2x-1}-1}{x-1}+\frac{\sqrt[5]{x-2}+1}{x-1}\)
Lại có \(\frac{\sqrt[4]{2x-1}-1}{x-1}=\frac{\sqrt{2x-1}-1}{\left(x-1\right)\left(\sqrt[4]{2x-1}+1\right)}=\frac{\left(2x-1\right)-1}{\left(x-1\right)\left(\sqrt[4]{2x-1}+1\right)\left(\sqrt{2x-1}+1\right)}=\frac{2}{\left(\sqrt[4]{2x-1}+1\right)\left(\sqrt{2x-1}+1\right)}\)
và đặt \(t=\sqrt[5]{x-2}\) thì khi \(x\rightarrow1\) có \(t\rightarrow-1,\) đồng thời \(\frac{\sqrt[5]{x-2}+1}{x-1}=\frac{t+1}{t^5+1}=\frac{1}{t^4-t^3+t^2-t+1}\) . Giới hạn cần tính bằng
\(\lim\limits_{x\rightarrow1}\frac{2}{\left(\sqrt[4]{2x-1}+1\right)\left(\sqrt{2x-1}+1\right)}+\lim\limits_{t\rightarrow-1}\frac{1}{t^4-t^3+t^2-y+1}=\frac{2}{4}+\frac{1}{5}=\frac{7}{10}.\)
