Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 1} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - 1 - {x^2}}}{{\sqrt {{x^2} - 1} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1}}{{\sqrt {{x^2} - 1} + x}}\\
= \frac{{ - 1}}{{ + \infty }} = 0\\
c,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {x - 4 - \sqrt {{x^2} - 7x + 2} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\left( {x - 4} \right)}^2} - \left( {{x^2} - 7x + 2} \right)}}{{x - 4 + \sqrt {{x^2} - 7x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - 8x + 16 - {x^2} + 7x - 2}}{{x - 4 + \sqrt {{x^2} - 7x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x + 14}}{{x - 4 + \sqrt {{x^2} - 7x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1 + \frac{{14}}{x}}}{{1 - \frac{4}{x} + \sqrt {1 - \frac{7}{x} + \frac{2}{{{x^2}}}} }}\\
= \frac{{ - 1}}{{1 + \sqrt 1 }}\\
= - \frac{1}{2}\\
e,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 7x + 1} - \sqrt {{x^2} - 3x + 2} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^2} - 7x + 1} \right) - \left( {{x^2} - 3x + 2} \right)}}{{\sqrt {{x^2} - 7x + 1} + \sqrt {{x^2} - 3x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 4x - 1}}{{\sqrt {{x^2} - 7x + 1} + \sqrt {{x^2} - 3x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 4 - \frac{1}{x}}}{{\sqrt {1 - \frac{7}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{3}{x} + \frac{2}{{{x^2}}}} }}\\
= \frac{{ - 4}}{{\sqrt 1 + \sqrt 1 }}\\
= - 2\\
f,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 4x + 1} - \sqrt {{x^2} - 9x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^2} - 4x + 1} \right) - \left( {{x^2} - 9x} \right)}}{{\sqrt {{x^2} - 4x + 1} + \sqrt {{x^2} - 9x} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{5x + 1}}{{\sqrt {{x^2} - 4x + 1} + \sqrt {{x^2} - 9x} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{5 + \frac{1}{x}}}{{\sqrt {1 - \frac{4}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{9}{x}} }}\\
= \frac{5}{{\sqrt 1 + \sqrt 1 }}\\
= \frac{5}{2}\\
g,\\
\mathop {\lim }\limits_{x \to + \infty } \sqrt x .\left( {\sqrt {x + 3} - \sqrt {x - 1} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \sqrt x .\frac{{\left( {x + 3} \right) - \left( {x - 1} \right)}}{{\sqrt {x + 3} + \sqrt {x - 1} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{4\sqrt x }}{{\sqrt {x + 3} + \sqrt {x - 1} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{4}{{\sqrt {1 + \frac{3}{x}} + \sqrt {1 - \frac{1}{x}} }}\\
= \frac{4}{{\sqrt 1 + \sqrt 1 }}\\
= 2
\end{array}\)
