Question

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 ___________.

Hint:

Expand the determinant to \(-(p^3+q^3)\) with \(p=a+\sin\tfrac x2,\ q=-b\cos x\). Existence of the limit forces \(a=b=0\), then use \(\sin\tfrac x2\sim\tfrac x2\).

Answer 1

Correct Answer: 14

Step 1: Let \(p=a+\sin\tfrac x2\), \(q=-b\cos x\). Expanding along the first row: \(F=p(0\cdot q-p\cdot p)-q(q\cdot q-0)=-(p^3+q^3)\).
Step 2: \(p^3+q^3=(p+q)(p^2-pq+q^2)\). As \(x\to0\): \(p\to a\), \(q\to-b\), so \(p^2-pq+q^2\to a^2+ab+b^2\), which is positive unless \(a=b=0\). Also \(p+q=(a-b)+\tfrac x2+\tfrac{b x^2}{2}+\dots\) has a nonzero \(x\)-term.
Step 3: If \(a^2+ab+b^2\neq0\), \(F\) behaves like a nonzero multiple of \(x\) (or a constant), so \(F/x^3\) cannot have a finite limit. Hence \(a=b=0\).
Step 4: Then \(F=-\sin^3\tfrac x2\), so \(L=\lim\dfrac{-\sin^3(x/2)}{x^3}=-\dfrac18\).
Step 5: \(-112L=-112\times\left(-\tfrac18\right)=14\).
Answer: 14