If `f (x)= |{:(x cos x, 2x sin x, x tan x),(1,x,1),(1,2x,1):}|,` find `lim _(xto0) (f(x))/(x ^(2)).`

Options

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

Answer

To solve the problem, we need to evaluate 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 \( f(x) \)

We can use the determinant formula for a 3x3 matrix:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

- \( a = x \cos x \)
- \( b = 2x \sin x \)
- \( c = x \tan x \)
- \( d = 1 \)
- \( e = x \)
- \( f = 1 \)
- \( g = 1 \)
- \( h = 2x \)
- \( i = 1 \)

Now, substituting these values into the determinant formula:

\[ f(x) = x \cos x \left( x \cdot 1 - 1 \cdot 2x \right) - 2x \sin x \left( 1 \cdot 1 - 1 \cdot 1 \right) + x \tan x \left( 1 \cdot 2x - 1 \cdot x \right) \]

### Step 2: Simplify the determinant

Calculating each term:

1. The first term:
\[ x \cos x (x - 2x) = x \cos x (-x) = -x^2 \cos x \]

2. The second term:
\[ -2x \sin x (1 - 1) = 0 \]

3. The third term:
\[ x \tan x (2x - x) = x \tan x (x) = x^2 \tan x \]

Combining these results:

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

### Step 3: Factor out \( x^2 \)

We can factor \( x^2 \) out of \( f(x) \):

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

### Step 4: Substitute into the limit

Now we substitute \( f(x) \) into the limit:

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

### Step 5: Evaluate the limit

Now we evaluate the limit:

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

At \( x = 0 \):

- \( \tan(0) = 0 \)
- \( \cos(0) = 1 \)

Thus:

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

### Final Result

Therefore, the limit is:

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