Question

\(\lim\limits_{x\rightarrow1}\dfrac{5}{\left(x-1\right)\left(x^2-3x+2\right)}\)

Answer 1

\(\lim\limits_{x\rightarrow1}\dfrac{5}{\left(x-1\right)\left(x^2-3x+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5}{x^3-3x^2+2x-x^2+3x-2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5}{x^3-4x^2+5x-2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{5}{x^3}}{1-\dfrac{4}{x}+\dfrac{5}{x^2}-\dfrac{2}{x^3}}=0\)

Answer 2

\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1}5=5>0\\\lim\limits_{x\rightarrow1}\left(x-1\right)\left(x^2-3x+2\right)=0\\\left(x-1\right)\left(x^2-3x+2\right)=\left(x-2\right)\left(x-1\right)^2< 0\end{matrix}\right.\) 

Suy ra: \(\lim\limits_{x\rightarrow1}\dfrac{5}{\left(x-1\right)\left(x^2-3x+2\right)}=-\infty\)