Nhận xét
\mathop {\lim }\limits_{x \to 1}[\frac{x}{x-1}-\frac{2}{x^{2}-1}] có dạng
\infty-\infty Ta có
\mathop {\lim }\limits_{x \to 1}[\frac{x}{x-1}-\frac{2}{x^{2}-1}]=\mathop {\lim }\limits_{x \to 1}\frac{x(x+1)-2}{x^{2}-1}
=\mathop {\lim }\limits_{x \to 1}\frac{x^{2}+x-2}{x^{2}-1} (Dạng \frac{0}{0})
=\mathop {\lim }\limits_{x\to 1}\frac{\left ( x-1 \right )\left ( x+2 \right )}{\left ( x-1 \right )\left ( x+1 \right )}
=\mathop {\lim }\limits_{x \to 1}\frac{x+2}{x+1}=\frac{3}{2}