Bài 6c (SGK-T133):Tính:\[\mathop {\lim }\limits_{x \to – \infty } (\sqrt {{x^2} – 2x + 5} )\]

$\mathop {\lim }\limits_{x \to – \infty } (\sqrt {{x^2} – 2x + 5} ) =
\mathop {\lim }\limits_{x \to – \infty } |x|\sqrt {1 – \frac{2}{x} + \frac{5}{{{x^2}}}} = + \infty $