We are given,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + {x^3}}  – \sqrt {1 – {x^3}} }}{{{x^2}}}$

Rationalize the numerator to get,

$ = \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\sqrt {1 + {{\text{x}}^3}}  – \sqrt {1 – {{\text{x}}^3}} }}{{{{\text{x}}^2}}} \times \left( {\frac{{\sqrt {1 + {{\text{x}}^3}}  + \sqrt {1 – {{\text{x}}^3}} }}{{\sqrt {1 + {{\text{x}}^3}}  + \sqrt {1 – {{\text{x}}^3}} }}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + {x^3}} \right) – \left( {1 – {x^3}} \right)}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 – {x^3}} } \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^3}}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 – {x^2}} } \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{\left( {\sqrt {1 + {x^3}}  + \sqrt {1 – {x^3}} } \right)}}$

Substituting the limit value to get,

$\mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{\left( {\sqrt {1 + {x^3}}  + \sqrt {1 – {x^3}} } \right)}} = 0$

Therefore, $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + {x^3}}  – \sqrt {1 – {x^3}} }}{{{x^2}}} = 0$.