Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{{x^2} - 1}}{{x - 1}}\,\,\,\,\,\,n\^e 'u\,\,\,\,\,x \ne 1\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n\^e 'u\,\,\,\,\,x = 1.\end{array} \right.\)

Tìm giới hạn \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) và so sánh giá trị này với f(1).

Lời giải:

Ta có: f(1) = 2.

\(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\)\( = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}}\)\( = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = 1 + 1 = 2\).

Vậy \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) = f(1).