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a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{2{n^2} - 3n + 1}}{{{n^2} + 1 \Rightarrow }}\). b) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2}}{{x - 1}}\) c) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 3} - 2x}}{{2x + 5}}\)

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Giải thích

a) \[\mathop {\lim }\limits_{n \to + \infty } \frac{{2{n^2} - 3n + 1}}{{{n^2} + 1}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{2 - \frac{3}{n} + \frac{1}{{{n^2}}}}}{{1 + \frac{1}{{{n^2}}}}} = 2.\]

b) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {x + 3} - 2} \right)\left( {\sqrt {x + 3} + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {\sqrt {x + 3} + 2} \right)}} = \frac{1}{4}.\)

c) \[\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 3} - 2x}}{{2x + 5}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {1 + \frac{3}{{{x^2}}}} - 2x}}{{2x + 5}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{3}{{{x^2}}}} - 2}}{{2 + \frac{5}{x}}} = - \frac{3}{2}.\]