giúp mik với
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{3x^2-6x+1}-x\sqrt{3}\right)=\lim\limits_{x\rightarrow-\infty}\left(\left|x\right|\sqrt{3-\dfrac{6}{x}+\dfrac{1}{x^2}}-x\sqrt{3}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(-x\sqrt{3-\dfrac{6}{x}+\dfrac{1}{x^2}}-x\sqrt{3}\right)=\lim\limits_{x\rightarrow-\infty}x\left(-\sqrt{3-\dfrac{6}{x}+\dfrac{1}{x^2}}-\sqrt{3}\right)\)
\(=-\infty.\left(-2\sqrt{3}\right)=+\infty\)
b.
\(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{3x^2+2x-1}{x+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(3x-1\right)}{x+1}\)
\(=\lim\limits_{x\rightarrow-1}\left(3x-1\right)=-4\)
\(f\left(-1\right)=\left(-1\right)^2-5=-4\)
\(\Rightarrow\lim\limits_{x\rightarrow-1}f\left(x\right)=f\left(-1\right)\Rightarrow\) hàm liên tục tại \(x=-1\)
