Do giới hạn \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}=a\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\)
\(\Rightarrow f\left(2\right)+1=0\Rightarrow f\left(2\right)=-1\)
\(T=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-2-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{f\left(x\right)+1+2\left(x-2\right)}{\sqrt{f\left(x\right)+2x+1}+2}-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{\dfrac{f\left(x\right)+1}{x-2}+2}{\sqrt{f\left(x\right)+2x+1}+2}-1}{x+2}\)
\(=\dfrac{\dfrac{a+2}{\sqrt{-1+4+1}+2}-1}{4}=\dfrac{a-2}{16}\)
