Suppose $\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$ exists and is equal to L, where

$$F(x) = \left| {\matrix{ {a + \sin {x \over 2}} & { - b\cos x} & 0 \cr { - b\cos x} & 0 & {a + \sin {x \over 2}} \cr 0 & {a + \sin {x \over 2}} & { - b\cos x} \cr } } \right|$$.

Then, $-$112 L is equal to ___________.

<p>Given,</p> <p>$$F(x) = \left| {\matrix{ {a + \sin {x \over 2}} & { - b\cos x} & 0 \cr { - b\cos x} & 0 & {a + \sin {x \over 2}} \cr 0 & {a + \sin {x \over 2}} & { - b\cos x} \cr } } \right|$$</p> <p>$$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$$</p> <p>$= - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$</p> <p>Now,</p> <p>$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$$</p> <p>Given limit exists, it only possible when a = 0 and b = 0.</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$$</p> <p>$= - {1 \over 8} \times 1 = L$</p> <p>$\therefore$ $- 112L = - 112 \times - {1 \over 8} = 14$</p>