a. Dạng 0.∞
Với \(x > -1\) đủ gần -1 (\(-1 < x < 0\)) ta có :
\(\eqalign{
& \left( {{x^3} + 1} \right)\sqrt {{x \over {{x^2} - 1}}} \cr &= \left( {{x^2} - x + 1} \right)\left( {x + 1} \right).\sqrt {{x \over {{x^2} - 1}}} \cr
& = \left( {{x^2} - x + 1} \right)\sqrt {{{x\left( {x + 1} \right)} \over {x - 1}}} \cr
& \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {{x \over {{x^2} - 1}}}\cr & \;\;= \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^2} - x + 1} \right)\sqrt {{{x\left( {x + 1} \right)} \over {x - 1}}} = 0 \cr} \)
b. Dạng 0.∞
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {{{x - 1} \over {{x^3} + x}}} \cr &= \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{{\left( {x + 2} \right)}^2}\left( {x - 1} \right)} \over {{x^3} + x}}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{{\left( {1 + {2 \over x}} \right)}^2}\left( {1 - {1 \over x}} \right)} \over {1 + {1 \over {{x^2}}}}}} = 1 \cr} \)
