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}}.
Sử dụng định lí về phép toán trên giới hạn hữu hạn của hàm số
Nếu \mathop {\lim }\limits_{x \to {x_0}} f(x) = L và \mathop {\lim }\limits_{x \to {x_0}} g(x) = M\left( {L,M \in \mathbb{R}} \right) thì
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\mathop {\lim }\limits_{x \to {x_0}} \left[ {f(x) \pm g(x)} \right] = L \pm M
\mathop {\lim }\limits_{x \to {x_0}} \left[ {f(x).g(x)} \right] = L.M
\mathop {\lim }\limits_{x \to {x_0}} \left[ {\frac{{f(x)}}{{g(x)}}} \right] = \frac{L}{M}\left( {M \ne 0} \right)
Nếu f(x) \ge 0 với mọi x \in \left( {a;b} \right)\backslash \left\{ {{x_0}} \right\} và \mathop {\lim }\limits_{x \to {x_0}} f(x) = L thì L \ge 0 và \mathop {\lim }\limits_{x \to {x_0}} \sqrt {f(x)} = \sqrt L .
a) \mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {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) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 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}{{\sqrt 1 + 1}} = \frac{1}{2}