Question

\(lim_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\left(\sqrt{\dfrac{3x}{x^2-1}}\right)\)

Answer 1

\(\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\cdot\sqrt{\dfrac{3x}{x^2-1}}\)

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

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

\(=\sqrt{-1+1}\left[\left(-1\right)^2-\left(-1\right)+1\right]\cdot\sqrt{\dfrac{3\left(-1\right)}{-1-2}}\)

=0