Question

Answer 1

\(lim\left(x->0\right)\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}=lim\left(x->0\right)\dfrac{\sqrt{4x+1}-1}{x}-\dfrac{\sqrt[3]{2x+1}-1}{x}=lim\left(x->0\right)\dfrac{4}{\sqrt{4x+1}+1}-\dfrac{2}{\sqrt[3]{2x+1}^2+\sqrt[3]{2x+1}+1}\\ =\dfrac{4}{\sqrt{1}+1}-\dfrac{2}{\sqrt[3]{1}^2+\sqrt[3]{1}+1}=\dfrac{4}{3}\)