Question

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

Answer 1

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

\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^-}\sqrt{2-x}=1>0\\\lim\limits_{x\rightarrow1^-}\left(4-x\right)\sqrt{1-x}=0\\x< 1\rightarrow\left(4-x\right)\sqrt{1-x}>0\end{matrix}\right.\\ \rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}=+\infty\)