Question

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)

Answer 1

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