<p>Let <span>\( f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\sin t & t & 2t \\ \sin t & t & t \end{vmatrix} \)</span>. Then find <span>\( \lim_{t \to 0} \dfrac{f(t)}{t^2} \)</span>.</p>
Step-by-Step Solution
Key Concept: We need to expand the determinant and analyze the behavior of f(t) as t → 0 using Taylor series or L'Hôpital's rule. Since f(t) is a determinant with entries involving trigonometric and polynomial functions, we must carefully track powers of t to evaluate the limit of f(t)/t².
Step 1: Expand the determinant using the first row.
$$f(t) = \cos t \begin{vmatrix} t & 2t \\ t & t \end{vmatrix} - t \begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix} + 1 \begin{vmatrix} 2\sin t & t \\ \sin t & t \end{vmatrix}$$
Step 2: Calculate the three $2 \times 2$ minors.
The minor for $\cos t$ is:
$$M_{11} = t \cdot t - 2t \cdot t = t^2 - 2t^2 = -t^2$$
The minor for $-t$ is:
$$M_{12} = 2\sin t \cdot t - 2t \cdot \sin t = 2t\sin t - 2t\sin t = 0$$
The minor for $1$ is:
$$M_{13} = 2\sin t \cdot t - t \cdot \sin t = 2t\sin t - t\sin t = t\sin t$$
Step 3: Substitute the minors back into the expansion for $f(t)$.
$$f(t) = \cos t (-t^2) - t (0) + 1 (t\sin t)$$
$$f(t) = -t^2\cos t + t\sin t$$
Step 4: Evaluate the limit.
$$\lim_{t \to 0} \frac{f(t)}{t^2} = \lim_{t \to 0} \frac{-t^2\cos t + t\sin t}{t^2}$$
$$= \lim_{t \to 0} \left( \frac{-t^2\cos t}{t^2} + \frac{t\sin t}{t^2} \right)$$
$$= \lim_{t \to 0} \left( -\cos t + \frac{\sin t}{t} \right)$$
Using the standard limits $\lim_{t \to 0} \cos t = 1$ and $\lim_{t \to 0} \frac{\sin t}{t} = 1$:
$$= -1 + 1 = 0$$
Correct Answer: 0