Question

Let \[ f(x)= \begin{vmatrix} -\cos x & \tan x & 3\sin x\\ 1 & -3x & 2x^2\\ x^3 & x & x^2 \end{vmatrix}. \] Then the value of \[ \lim_{x\to 0}\frac{(1+x)f(x)-3x\sin x}{x^3} \] is:

• A. \(3\)
• B. \(5\)
• C. \(7\)
• D. \(8\)

Hint:

When limits involve determinants and small \(x\), expand both the determinant and trigonometric functions only up to the necessary order.

Answer 1

The Correct Option is C

Concept:
Expand the determinant and retain terms up to the required order of \(x\).
Use standard Taylor expansions of trigonometric functions near \(x=0\).
Since the limit involves division by \(x^3\), terms beyond \(x^3\) can be ignored.
Step 1: Expand the determinant \[ f(x)= -\cos x \begin{vmatrix} -3x & 2x^2\\ x & x^2 \end{vmatrix} -\tan x \begin{vmatrix} 1 & 2x^2\\ x^3 & x^2 \end{vmatrix} +3\sin x \begin{vmatrix} 1 & -3x\\ x^3 & x \end{vmatrix} \] \[ =5x^3\cos x -x^2\tan x +3x\sin x + O(x^4) \]
Step 2: Use Taylor expansions \[ \cos x = 1-\frac{x^2}{2}, \quad \tan x = x+\frac{x^3}{3}, \quad \sin x = x-\frac{x^3}{6} \] Substituting, \[ f(x)=3x^2+4x^3+O(x^4) \]
Step 3: Evaluate the numerator \[ (1+x)f(x)=3x^2+7x^3+O(x^4) \] \[ 3x\sin x=3x^2+O(x^4) \] \[ (1+x)f(x)-3x\sin x=7x^3+O(x^4) \]
Step 4: Take the limit \[ \lim_{x\to 0}\frac{(1+x)f(x)-3x\sin x}{x^3} =\lim_{x\to 0}\frac{7x^3}{x^3}=7 \]