Question

Let \[ f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{vmatrix}, \] then \[ \lim_{x \to 0} \frac{f(x)}{x^2} = ? \]

• A. 2
• B. -2
• C. 1
• D. -1

Hint:

When faced with limits that result in indeterminate forms like 0/0 , use L’Hopital’s Rule to differentiate the numerator and denominator until a solvable limit is found.

Answer 1

The Correct Option is B

The determinant of f(x) is:

\[ f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{vmatrix}. \]

Step 1: Expand the determinant.

Expand along the first row:

\[ f(x) = \cos x \cdot \begin{vmatrix}x^3 & 2x \\ x & 1 \end{vmatrix} - x \cdot \begin{vmatrix}2 \sin x & 2x \\ \tan x & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix}2 \sin x & x^3 \\ \tan x & x \end{vmatrix}. \]

Simplify each minor determinant:

1. \[ \begin{vmatrix}x^3 & 2x \\ x & 1 \end{vmatrix} = x^3 \cdot 1 - 2x \cdot x = x^3 - 2x^2. \]
2. \[ \begin{vmatrix}2 \sin x & 2x \\ \tan x & 1 \end{vmatrix} = (2 \sin x)(1) - (2x)(\tan x) = 2 \sin x - 2x \tan x. \]
3. \[ \begin{vmatrix}2 \sin x & x^3 \\ \tan x & x \end{vmatrix} = (2 \sin x)(x) - (x^3)(\tan x) = 2x \sin x - x^3 \tan x. \]

Thus:

\[ f(x) = \cos x (x^3 - 2x^2) - x (2 \sin x - 2x \tan x) + (2x \sin x - x^3 \tan x). \]

Step 2: Simplify \(\frac{f(x)}{x^2}\).

Divide \(f(x)\) by \(x^2\):

\[ \frac{f(x)}{x^2} = \frac{\cos x (x^3 - 2x^2)}{x^2} - \frac{x (2 \sin x - 2x \tan x)}{x^2} + \frac{2x \sin x - x^3 \tan x}{x^2}. \]

Simplify each term:

1. First term:
2. Second term:
3. Third term:

Thus:

\[ \frac{f(x)}{x^2} = \cos x (x - 2) - \left(\frac{2 \sin x}{x} - 2 \tan x\right) + \left(\frac{2 \sin x}{x} - x^2 \tan x\right). \]

Step 3: Take the limit as \(x \to 0\).

Using standard limits:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \tan x = x, \quad \lim_{x \to 0} \cos x = 1, \]

substitute \(x \to 0\):

1. First term:
2. Second term:
3. Third term:

Combine terms:

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

Conclusion: The value of \(\lim_{x \to 0} \frac{f(x)}{x^2}\) is:

\[ \boxed{-2}. \]