Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
18,\\
\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - {a^2}x + 1} - x - 1} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - {a^2}x + 1} - \left( {x + 1} \right)} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {{x^2} - {a^2}x + 1} - \left( {x + 1} \right)} \right)\left( {\sqrt {{x^2} - {a^2}x + 1} + \left( {x + 1} \right)} \right)}}{{\sqrt {{x^2} - {a^2}x + 1} + \left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^2} - {a^2}x + 1} \right) - {{\left( {x + 1} \right)}^2}}}{{\sqrt {{x^2} - {a^2}x + 1} + \left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - \left( {{a^2} + 2} \right)x}}{{\sqrt {{x^2} - {a^2}x + 1} + \left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - {a^2} + 2}}{{\sqrt {1 - \frac{{{a^2}}}{x} + \frac{1}{{{x^2}}}} + 1 + \frac{1}{x}}}\\
= \frac{{ - {a^2} + 2}}{{\sqrt 1 + 1}}\\
= \frac{{ - {a^2}}}{2} + 1\\
20,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {a{x^2} + x + 1} - \sqrt {{x^2} + bx - 2} } \right) = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {a{x^2} + x + 1} - \sqrt {{x^2} + bx - 2} } \right)\left( {\sqrt {a{x^2} + x + 1} + \sqrt {{x^2} + bx - 2} } \right)}}{{\sqrt {a{x^2} + x + 1} + \sqrt {{x^2} + bx - 2} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {a{x^2} + x + 1} \right) - \left( {{x^2} + bx - 2} \right)}}{{\sqrt {a{x^2} + x + 1} + \sqrt {{x^2} + bx - 2} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {a - 1} \right){x^2} + \left( {1 - b} \right)x + 3}}{{\sqrt {a{x^2} + x + 1} + \sqrt {{x^2} + bx - 2} }} = 2\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {a - 1} \right)x + \left( {1 - b} \right) + \frac{3}{x}}}{{\sqrt {a + \frac{1}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 + \frac{b}{x} - \frac{2}{x}} }} = 2\\
\Rightarrow \left\{ \begin{array}{l}
a - 1 = 0\\
\frac{{1 - b}}{{\sqrt a + \sqrt 1 }} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = - 3
\end{array} \right. \Rightarrow P = - 3
\end{array}\)
