a. Ta có: \({{{x^3} + 3\sqrt 3 } \over {3 - {x^2}}} = {{\left( {x + \sqrt 3 } \right)\left( {{x^2} - x\sqrt 3 + 3} \right)} \over {\left( {x + \sqrt 3 } \right)\left( {\sqrt 3 - x} \right)}} = {{{x^2} - x\sqrt 3 + 3} \over {\sqrt 3 - x}}\)
với \(\,x \ne - \sqrt 3 \)
Do đó : \(\mathop {\lim }\limits_{x \to - \sqrt 3 } {{{x^3} + 3\sqrt 3 } \over {3 - {x^2}}} =\mathop {\lim }\limits_{x \to - \sqrt 3 } {{{x^2} - x\sqrt 3 + 3} \over {\sqrt 3 - x}}= {9 \over {2\sqrt 3 }} = {{3\sqrt 3 } \over 2}\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 4} {{\sqrt x - 2} \over {{x^2} - 4x}} = \mathop {\lim }\limits_{x \to 4} {{\sqrt x - 2} \over {x\left( {x - 4} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 4} {1 \over {x\left( {\sqrt x + 2} \right)}} = {1 \over {16}} \cr} \)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x - 1} } \over {{x^2} - x}} = \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x - 1} } \over {x\left( {x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} {1 \over {x\sqrt {x - 1} }} = + \infty \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + x + 1} - 1} \over {3x}} = \mathop {\lim }\limits_{x \to 0} {{{x^2} + x + 1 - 1} \over {3x(\sqrt {{x^2} + x + 1} + 1)}} \cr
& = {1 \over 3}\mathop {\lim }\limits_{x \to 0} {{x + 1} \over {\sqrt {{x^2} + x + 1} + 1}} = {1 \over 6} \cr} \)
