Official solution
Given limit \[ = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}\]
\[ = \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}\]
\[ = \mathop {\lim }\limits_{x \to \alpha } {2 \over {{{\left( {x - \alpha } \right)}^2}}} \times {{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}}} \times {{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}\]
\[ = {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.\]