Question

If \[ f(x)= \begin{vmatrix} 2+\dfrac{\sin x}{x} & 1 & 3
[4pt] 2 & 1+\dfrac{\sin x}{x} & 3
[4pt] 2 & 1 & 3+\dfrac{\sin x}{x} \end{vmatrix}, \] then \[ \lim_{x\to 0} f(x)= \]

• A. \(1\)
• B. \(0\)
• C. \(5\)
• D. \(7\) \bigskip

Hint:

Whenever a determinant contains \[ \frac{\sin x}{x}, \] first use \[ \lim_{x\to0}\frac{\sin x}{x}=1, \] then evaluate the resulting determinant directly.

Answer 1

The Correct Option is D

Concept: Use \[ \lim_{x\to0}\frac{\sin x}{x}=1. \] Since the determinant is a polynomial expression in its entries, the limit can be evaluated by direct substitution.

Step 1: Apply the standard limit. Let \[ t=\frac{\sin x}{x}. \] As \[ x\to0, \] \[ t\to1. \] Therefore, \[ \lim_{x\to0}f(x) = \begin{vmatrix} 3&1&3 2&2&3 2&1&4 \end{vmatrix}. \]

Step 2: Evaluate the determinant. Expanding along the first row, \[ \begin{vmatrix} 3&1&3 2&2&3 2&1&4 \end{vmatrix} = 3 \begin{vmatrix} 2&3 1&4 \end{vmatrix} - 1 \begin{vmatrix} 2&3 2&4 \end{vmatrix} + 3 \begin{vmatrix} 2&2 2&1 \end{vmatrix}. \] \[ = 3(8-3) -(8-6) +3(2-4). \] \[ = 15-2-6. \] \[ =7. \]

Step 3: Write the final answer. \[ \boxed{7} \]