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