a)\(\mathop {\lim }\limits_{x \to + \infty } ({x^4} - {x^2} + x - 1) \)

\(= \mathop {\lim }\limits_{x \to + \infty } {x^4}\left( {1 - {1 \over {{x^2}}} + {1 \over {{x^3}}} - {1 \over {{x^4}}}} \right) \)

\(= \mathop {\lim }\limits_{x \to + \infty } {x^4}.\mathop {\lim }\limits_{x \to + \infty }\left( {1 - {1 \over {{x^2}}} + {1 \over {{x^3}}} - {1 \over {{x^4}}}} \right) =+ \infty \)

Vì \(\mathop {\lim }\limits_{x \to + \infty } {x^4}=+\infty \) (giới hạn đặc biệt của hàm số)

\(\mathop {\lim }\limits_{x \to + \infty }\left( {1 - {1 \over {{x^2}}} + {1 \over {{x^3}}} - {1 \over {{x^4}}}} \right)=1>0\)

Giá trị $(+)$ nhân $(+)$ sẽ bằng $(+)$

b)\(\mathop {\lim }\limits_{x \to - \infty } ( - 2{x^3} + 3{x^2} - 5) \)

\(= \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 2 + {1 \over x} - {5 \over {{x^2}}}} \right)\)

\(=\mathop {\lim }\limits_{x \to - \infty } {x^3}.\mathop {\lim }\limits_{x \to - \infty }\left( { - 2 + {1 \over x} - {5 \over {{x^2}}}} \right) = + \infty \)

Vì \(\mathop {\lim }\limits_{x \to + \infty } {x^3}=-\infty \) (giới hạn đặc biệt của hàm số)

\(\mathop {\lim }\limits_{x \to - \infty }\left( { - 2 + {1 \over x} - {5 \over {{x^2}}}} \right)=-2<0\)

Giá trị $(-)$ nhân $(-)$ sẽ bằng $(+)$

c)\(\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} - 2x + 5} ) = \mathop {\lim }\limits_{x \to - \infty } |x|\sqrt {1 - {2 \over x} + {5 \over {{x^2}}}} = + \infty \)

d)\(\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} + 1} + x} \over {5 - 2x}}\)

\(= \mathop {\lim }\limits_{x \to + \infty } {{x\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {5 - 2x}}\)

\(= \mathop {\lim }\limits_{x \to + \infty } {{\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {{5 \over x} - 2}}\)

\(= {{\mathop {\lim }\limits_{x \to + \infty }\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {\mathop {\lim }\limits_{x \to + \infty }{5 \over x} - 2}}\)

\(=\frac{1+1}{-2}= - 1 \)