a. Dạng \({0 \over 0}\) ta phân tích tử và mẫu ra thừa số :
\(\eqalign{
& \mathop {\lim }\limits_{x \to 2} {{{x^3} - 8} \over {{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} {{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} \over {\left( {x - 2} \right)\left( {x + 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 2} {{{x^2} + 2x + 4} \over {x + 2}} = 3 \cr} \)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{\left( {x + 3} \right)\left( {2x - 1} \right)} \over {{{\left( {x + 3} \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{2x - 1} \over {x + 3}} = - \infty \cr} \)
Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \left( {2x - 1} \right) = - 7 < 0\,\text{ và }\,\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ +}} \left( {x + 3} \right) = 0;\)
\(x + 3 > 0\)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{\left( {x + 3} \right)\left( {2x - 1} \right)} \over {{{\left( {x + 3} \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{2x - 1} \over {x + 3}} = + \infty \cr} \)
Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {2x - 1} \right) = - 7 < 0\,\text{ và }\,\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {x + 3} \right) = 0;\)
\(x + 3 < 0\)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^3} + 1} - 1} \over {{x^2} + x}} = \mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {x\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1} + 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} {{{x^2}} \over {\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1} + 1} \right)}} = 0 \cr} \)
