`f'(x) = |{:(cos x , x , 1),(2sin x,x^(2),2x),(tan x, x,1):}|` . The value of `underset(x rarr 0)(lim) (f(x))/(x)` is equal to

Options

A. 1
B. `-1`
C. 0
D. None of these

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{f(x)}{x} \]

where \( f'(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right| \).

### Step 1: Calculate the Determinant

We need to compute the determinant \( f'(x) \):

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

Using the formula for the determinant of a \( 3 \times 3 \) matrix, we can expand it:

\[ f'(x) = \cos x \left| \begin{array}{cc} x^2 & 2x \\ x & 1 \end{array} \right| - x \left| \begin{array}{cc} 2 \sin x & 2x \\ \tan x & 1 \end{array} \right| + 1 \left| \begin{array}{cc} 2 \sin x & x^2 \\ \tan x & x \end{array} \right| \]

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

\[ \left| \begin{array}{cc} x^2 & 2x \\ x & 1 \end{array} \right| = x^2 \cdot 1 - 2x \cdot x = x^2 - 2x^2 = -x^2 \]

2. For the second determinant:

\[ \left| \begin{array}{cc} 2 \sin x & 2x \\ \tan x & 1 \end{array} \right| = 2 \sin x \cdot 1 - 2x \cdot \tan x = 2 \sin x - 2x \tan x \]

3. For the third determinant:

\[ \left| \begin{array}{cc} 2 \sin x & x^2 \\ \tan x & x \end{array} \right| = 2 \sin x \cdot x - x^2 \cdot \tan x = 2x \sin x - x^2 \tan x \]

### Step 3: Substitute Back into the Determinant

Now substituting these back into the expression for \( f'(x) \):

\[ f'(x) = \cos x (-x^2) - x (2 \sin x - 2x \tan x) + (2x \sin x - x^2 \tan x) \]

### Step 4: Simplify the Expression

Simplifying the expression:

\[ f'(x) = -x^2 \cos x - 2x \sin x + 2x^2 \tan x + 2x \sin x - x^2 \tan x \]

The \( -2x \sin x + 2x \sin x \) cancels out:

\[ f'(x) = -x^2 \cos x + x^2 \tan x \]

### Step 5: Factor Out \( x^2 \)

Factoring out \( x^2 \):

\[ f'(x) = x^2 (\tan x - \cos x) \]

### Step 6: Find \( f(x) \)

To find \( f(x) \), we need to integrate \( f'(x) \):

\[ f(x) = \int f'(x) \, dx \]

However, for the limit we are interested in, we can directly evaluate:

### Step 7: Evaluate the Limit

Now, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{f'(x)}{x} \]

Substituting \( f'(x) = x^2 (\tan x - \cos x) \):

\[ \lim_{x \to 0} \frac{x^2 (\tan x - \cos x)}{x} = \lim_{x \to 0} x (\tan x - \cos x) \]

### Step 8: Use L'Hospital's Rule if Necessary

As \( x \to 0 \), both \( \tan x \) and \( \cos x \) approach 0. We can apply L'Hospital's Rule:

\[ \lim_{x \to 0} \frac{\tan x - \cos x}{1/x} = \lim_{x \to 0} \frac{\sec^2 x + \sin x}{-1/x^2} \]

Evaluating this limit yields 0 as \( x \to 0 \).

### Final Answer

Thus, the value of

\[ \lim_{x \to 0} \frac{f(x)}{x} = 0. \]