Question

Answer 1

\(lim\left(x->0\right)\dfrac{\sqrt[3]{x+1}-1}{\sqrt{2x+1}-1}=lim\left(x->0\right)\dfrac{\left(x+1\right)-1}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1}:\dfrac{2x+1-1}{\sqrt{2x+1}+1}=lim\left(x->0\right)\dfrac{\sqrt{2x+1}+1}{2\left(\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1\right)}\\ =\dfrac{\sqrt{2\cdot0+1}+1}{2\left(\sqrt[3]{0+1}^2+\sqrt[3]{0+1}+1\right)}=\dfrac{1}{3}\)