\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} - x + 1}}{{{x^2} - 1}}\) bằng:

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} - x + 1}}{{{x^2} - 1}} =  + \infty \)

vì \(\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - x + 1} \right) = 1 > 0,\mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} - 1} \right) = 0,{x^2} - 1 > 0,\forall x > 1\)