Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 2} \right)}^2} - 4}}{x}\);

b) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 9} - 3}}{{{x^2}}}\).

Lời giải:

a) Ta có: \(\mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 1} \right) = 0\), x – 1 > 0 với mọi x > 1 và

\(\mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 2} \right) = 1 - 2 = - 1 < 0\).

Do đó, \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 2}}{{x - 1}} = - \infty \).

b) Ta có: \(\mathop {\lim }\limits_{x \to {4^ - }} \left( {4 - x} \right) = 0\), 4 – x > 0 với mọi x < 4 và

\(\mathop {\lim }\limits_{x \to {4^ - }} \left( {{x^2} - x + 1} \right) = {4^2} - 4 + 1 = 13 > 0\).

Do đó, \(\mathop {\lim }\limits_{x \to {4^ - }} \frac{{{x^2} - x + 1}}{{4 - x}} = + \infty \).

Lời giải:

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

b) Ta có: \(\sqrt {{x^2} + x + 2} - x = \frac{{{{\left( {\sqrt {{x^2} + x + 2} } \right)}^2} - {x^2}}}{{\sqrt {{x^2} + x + 2} + x}}\)\( = \frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + x}}\)

Do đó, \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 2} - x} \right)\) \( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + x}}\)

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