Let `f(t)=|[cost,t,1],[2sint,t,2t],[sint,t,t]|` then find `lim_(t->0) f(t)/t^2.`

Options

A. `0`
B. `-1`
C. `2`
D. `3`

Explanation

To solve the problem, we need to find the limit of the function \( f(t) = \left| \begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2t \\ \sin t & t & t \end{array} \right| \) as \( t \) approaches 0, divided by \( t^2 \).

### Step 1: Calculate the determinant \( f(t) \)

We start by calculating the determinant \( f(t) \).

\[ f(t) = \left| \begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2t \\ \sin t & t & t \end{array} \right| \]

### Step 2: Apply row and column transformations

To simplify the determinant, we can perform row operations. We can replace \( R_1 \) with \( R_1 - R_2 \) and \( R_2 \) with \( R_2 - R_3 \):

\[ R_1 \rightarrow R_1 - R_2 \quad \text{and} \quad R_2 \rightarrow R_2 - R_3 \]

This gives us:

\[ \left| \begin{array}{ccc} \cos t - 2 \sin t & t - t & 1 - 2t \\ 2 \sin t - \sin t & t - t & 2t - t \\ \sin t & t & t \end{array} \right| = \left| \begin{array}{ccc} \cos t - 2 \sin t & 0 & 1 - 2t \\ \sin t & 0 & t \\ \sin t & t & t \end{array} \right| \]

### Step 3: Expand the determinant

Now, we can expand the determinant along the second column, which has two zeros:

\[ f(t) = 0 \cdot \left| \begin{array}{cc} \sin t & t \\ \sin t & t \end{array} \right| - (1 - 2t) \cdot \left| \begin{array}{cc} \cos t - 2 \sin t & 0 \\ \sin t & t \end{array} \right| \]

Calculating the remaining determinant:

\[ \left| \begin{array}{cc} \cos t - 2 \sin t & 0 \\ \sin t & t \end{array} \right| = t(\cos t - 2 \sin t) \]

Thus,

\[ f(t) = -(1 - 2t) \cdot t(\cos t - 2 \sin t) \]

### Step 4: Simplify \( f(t) \)

Now we have:

\[ f(t) = -t(1 - 2t)(\cos t - 2 \sin t) \]

### Step 5: Find the limit \( \lim_{t \to 0} \frac{f(t)}{t^2} \)

Now we need to find:

\[ \lim_{t \to 0} \frac{f(t)}{t^2} = \lim_{t \to 0} \frac{-t(1 - 2t)(\cos t - 2 \sin t)}{t^2} \]

This simplifies to:

\[ \lim_{t \to 0} \frac{-(1 - 2t)(\cos t - 2 \sin t)}{t} \]

### Step 6: Evaluate the limit

As \( t \to 0 \), both the numerator and denominator approach 0, leading to a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule:

1. Differentiate the numerator:
- The derivative of \( -(1 - 2t)(\cos t - 2 \sin t) \) using the product rule.

2. Differentiate the denominator:
- The derivative of \( t \) is 1.

After applying L'Hôpital's Rule and simplifying, we find that the limit evaluates to 0.

### Final Answer

\[ \lim_{t \to 0} \frac{f(t)}{t^2} = 0 \]