Đề bài

Mỗi giới hạn sau có tồn tại không? Nếu có, hãy tìm giới hạn đó. a) \(\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{\left| x \right|}}\); b) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}}\).

Lời giải chi tiết

a) Ta có: \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{{\left| x \right|}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} x = 0\); \(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^2}}}{{\left| x \right|}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^2}}}{{ - x}} = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0\) Vì \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{{\left| x \right|}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^2}}}{{\left| x \right|}} = 0\) nên \(\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{\left| x \right|}} = 0\); b) Ta có: \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x\left( {x - 2} \right)}}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^ - }} \left( { - x} \right) =  - 2\); \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{x\left( {x - 2} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} x = 2\) Vì \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}} \ne \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}}\) nên không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{\left| {x - 2} \right|}}\).