Question

\(\lim\limits_{x\rightarrow0^-}\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)\)

\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)

\(\lim\limits_{x\rightarrow1^+}\dfrac{1}{x^3-1}-\dfrac{1}{x-1}\)

\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt[3]{1-x^3}\right)\)

 

Answer 1

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v