Giải thích các bước giải:
\(\begin{array}{l}
b,\\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^4} - 5{x^2} + 1} \\
= \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^4}.\left( {1 - \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}}} \right)} \\
= \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^4}} .\sqrt {1 - \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}}} \\
= \mathop {\lim }\limits_{x \to - \infty } {x^2}.\sqrt {1 - \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}}} \\
\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2}} \right) = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {1 - \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}}} = \sqrt {1 - 5.0 + 0} = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - \infty } {x^2}.\sqrt {1 - \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}}} = + \infty \\
\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^4} - 5{x^2} + 1} = + \infty \\
c,\\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {3{x^2} - 2x + 5} \\
= \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2}.\left( {3 - \frac{2}{x} + \frac{5}{{{x^2}}}} \right)} \\
= \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2}} .\sqrt {3 - \frac{2}{x} + \frac{5}{{{x^2}}}} \\
= \mathop {\lim }\limits_{x \to - \infty } \left| x \right|.\sqrt {3 - \frac{2}{x} + \frac{5}{{{x^2}}}} \\
\mathop {\lim }\limits_{x \to - \infty } \left| x \right| = + \infty \\
\mathop {\lim }\limits_{x \to - \infty } \sqrt {3 - \frac{2}{x} + \frac{5}{{{x^2}}}} = 3\\
\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \sqrt {3{x^2} - 2x + 5} = + \infty
\end{array}\)
