Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{{x^3} + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^2} - x + 1} \right) = {\left( { - 1} \right)^2} - \left( { - 1} \right) + 1 = 3\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{\sqrt {x + 5} - 2}}{{\sqrt { - x} - 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{\frac{{x + 5 - {2^2}}}{{\sqrt {x + 5} + 2}}}}{{\frac{{ - x - 1}}{{\sqrt { - x} + 1}}}}\\
= \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \frac{{\frac{{x + 1}}{{\sqrt {x + 5} + 2}}}}{{\frac{{ - \left( {x + 1} \right)}}{{\sqrt { - x} + 1}}}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( { - \frac{{\sqrt { - x} + 1}}{{\sqrt {x + 5} + 2}}} \right)\\
= - \frac{{\sqrt 1 + 1}}{{\sqrt { - 1 + 5} + 2}} = - \frac{2}{4} = - \frac{1}{2}\\
\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right)
\end{array}\)
Do đó, không tồn tại \(\mathop {\lim }\limits_{x \to \left( { - 1} \right)} f\left( x \right)\)
