\(\lim\limits_{x\rightarrow1}\left[\dfrac{2}{\left(x-1\right)^2}\cdot\dfrac{2x+1}{2x-3}\right]\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1}\left(x-1\right)^2=\left(1-1\right)^2=0\\\lim\limits_{x\rightarrow1}\dfrac{2x+1}{2x-3}=\dfrac{2\cdot1+1}{2\cdot1-3}=\dfrac{3}{-1}=-3< 0\\\lim\limits_{x\rightarrow1}2=2>0\\\end{matrix}\right.\)