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a) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}}\) b) \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^3} - {x^2}} }}{{\sqrt {x - 1} + 1 - x}}\)

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Giải thích

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x + 4}}{{x + 2}} = \frac{{12}}{4} = 3\).

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