Đáp án:
\[\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 3x + 5} - x} \right) = \frac{{ - 3}}{2}\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 3x + 5} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {{x^2} - 3x + 5} - x} \right)\left( {\sqrt {{x^2} - 3x + 5} + x} \right)}}{{\sqrt {{x^2} - 3x + 5} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^2} - 3x + 5} \right) - {x^2}}}{{\sqrt {{x^2} - 3x + 5} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 3x + 5}}{{\sqrt {{x^2} - 3x + 5} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 3 + \frac{5}{x}}}{{\sqrt {1 - \frac{3}{x} + \frac{5}{{{x^2}}}} + 1}}\\
= \frac{{ - 3 + 0}}{{\sqrt {1 - 0 + 0} + 1}}\\
= \frac{{ - 3}}{2}
\end{array}\)
