Đáp án:
$TCĐ: x=-1\\TCN: y=1$
Giải thích các bước giải:
$ y=\dfrac{x^2-5x+4}{x^2-1}=\dfrac{(x-4)(x-1)}{(x-1)(x+1)}=\dfrac{x-4}{x+1}\\ \displaystyle\lim_{x \to -1^+} y\\ =\displaystyle\lim_{x \to -1^+} \dfrac{x-4}{x+1}\\ =-\infty\\ \Rightarrow TCĐ: x=-1\\ \displaystyle\lim_{x \to +\infty} y\\ =\displaystyle\lim_{x \to +\infty} \dfrac{x^2-5x+4}{x^2-1}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{1-\dfrac{5}{x}+\dfrac{4}{x^2}}{1-\dfrac{1}{x^2}}\\ =1\\ \Rightarrow TCN: y=1$
