\(A=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4+x+x^2}-2}{x+1}=\)\(\lim\limits_{x\rightarrow-1}\dfrac{\left|x\right|\sqrt{\dfrac{4}{x^2}+\dfrac{1}{x}+1}-2}{x\left(1+\dfrac{1}{x}\right)}\left(1\right)\)
\(TH1:x\ge0\)
\(\)\(\left(1\right)\Rightarrow A=\lim\limits_{x\rightarrow-1}\dfrac{x\sqrt{\dfrac{4}{x^2}+\dfrac{1}{x}+1}-2}{x\left(1+\dfrac{1}{x}\right)}=\)\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{\dfrac{4}{x^2}+\dfrac{1}{x}+1}-2}{1+\dfrac{1}{x}}=-\infty\)
\(TH2:x< 0\)
\(\left(1\right)\Rightarrow A=\lim\limits_{x\rightarrow-1}\dfrac{-x\sqrt{\dfrac{4}{x^2}+\dfrac{1}{x}+1}-2}{x\left(1+\dfrac{1}{x}\right)}=\)\(\lim\limits_{x\rightarrow-1}\dfrac{-\sqrt{\dfrac{4}{x^2}+\dfrac{1}{x}+1}-2}{1+\dfrac{1}{x}}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4+x+x^2}-2}{x+1}=\lim\limits_{x\rightarrow-1}\dfrac{x^2+x}{\left(x+1\right)\left(\sqrt{4+x+x^2}+2\right)}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{x}{\sqrt{4+x+x^2}+2}=\dfrac{-1}{\sqrt{4-1+1}+2}=-\dfrac{1}{4}\)
tính nguyên hàm của hàm số f(x)=