Let `f(x)=|(cosx,x,1),(2sinx,x^(2),2x),(tan x,x,1)|` then `lim_(xto0)(f(x))/(x^(2))` is given by

Options

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

Answer

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

### Step-by-Step Solution:

1. **Calculate the Determinant \( f(x) \)**:
We need to compute the determinant of the given matrix:
\[ f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right| \]

2. **Use the Determinant Formula**:
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:
- \( a = \cos x, b = x, c = 1 \)
- \( d = 2\sin x, e = x^2, f = 2x \)
- \( g = \tan x, h = x, i = 1 \)

Thus, we can expand the determinant.

3. **Expand the Determinant**:
\[ f(x) = \cos x \left( x^2 \cdot 1 - 2x \cdot x \right) - x \left( 2\sin x \cdot 1 - 2x \cdot \tan x \right) + 1 \left( 2\sin x \cdot x - x^2 \cdot \tan x \right) \]
Simplifying this gives:
\[ f(x) = \cos x (x^2 - 2x^2) - x (2\sin x - 2x \tan x) + (2x \sin x - x^2 \tan x) \]
\[ = -x^2 \cos x - x (2\sin x - 2x \tan x) + (2x \sin x - x^2 \tan x) \]

4. **Combine Like Terms**:
After simplifying, we can combine terms:
\[ = -x^2 \cos x + 2x \sin x - 2x^2 \tan x \]

5. **Evaluate the Limit**:
Now, we need to evaluate:
\[ \lim_{x \to 0} \frac{f(x)}{x^2} \]
Substituting \( f(x) \):
\[ = \lim_{x \to 0} \frac{-x^2 \cos x + 2x \sin x - 2x^2 \tan x}{x^2} \]
This simplifies to:
\[ = \lim_{x \to 0} \left(-\cos x + \frac{2\sin x}{x} - 2\tan x\right) \]

6. **Apply Limits**:
As \( x \to 0 \):
- \( \cos x \to 1 \)
- \( \frac{\sin x}{x} \to 1 \)
- \( \tan x \to 0 \)

Thus:
\[ = -1 + 2(1) - 0 = -1 + 2 = 1 \]

7. **Final Result**:
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{f(x)}{x^2} = -1 \]

### Final Answer:
The limit is \( -1 \).