Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right)\);
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}}\);
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}}\).
Lời giải
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = {2^2} - 4.2 + 3 = - 1\).
b) \[\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = 1\].
c) \[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{2}\].