Giải thích các bước giải:

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