if `f(x)=|[xcosx,2xsinx,xtanx],[1,x,1],[1,2x,1]| ` find `lim_(x->0) f(x)/x^2`

Options

A. 0
B. 1
C. `-1`
D. Does not exist

Answer

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

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

where

\[ f(x) = \begin{vmatrix} x \cos x & 2x \sin x & x \tan x \\ 1 & x & 1 \\ 1 & 2x & 1 \end{vmatrix} \]

### Step 1: Calculate the Determinant

First, we will compute the determinant \( f(x) \).

\[ f(x) = \begin{vmatrix} x \cos x & 2x \sin x & x \tan x \\ 1 & x & 1 \\ 1 & 2x & 1 \end{vmatrix} \]

### Step 2: Simplify the Determinant

To simplify the determinant, we can perform a row operation. We will subtract the second row from the third row:

\[ \text{Row 3} = \text{Row 3} - \text{Row 2} \]

This gives us:

\[ f(x) = \begin{vmatrix} x \cos x & 2x \sin x & x \tan x \\ 1 & x & 1 \\ 0 & x & 0 \end{vmatrix} \]

### Step 3: Expand the Determinant

Now we can expand the determinant along the first column:

\[ f(x) = x \cos x \cdot \begin{vmatrix} x & 1 \\ x & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2x \sin x & x \tan x \\ x & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} x & 1 \\ x & 0 \end{vmatrix} = (x)(0) - (1)(x) = -x \]

2. For the second determinant:
\[ \begin{vmatrix} 2x \sin x & x \tan x \\ x & 0 \end{vmatrix} = (2x \sin x)(0) - (x \tan x)(x) = -x^2 \tan x \]

Putting it all together:

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

### Step 4: Find the Limit

Now we need to find:

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

### Step 5: Evaluate the Limit

As \( x \to 0 \):

- \(\tan x \to 0\)
- \(\cos x \to 1\)

Thus,

\[ \lim_{x \to 0} (\tan x - \cos x) = 0 - 1 = -1 \]

### Final Answer

Therefore, the limit is:

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