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 \(\lim_{x\to0} \frac{\sin x}{x}=1\). As \(x\to0\), \(\frac{\sin x}{x}\to1\), so the determinant limit is the determinant with 1 substituted.

Step 1: \(\lim_{x\to0} f(x) = \begin{vmatrix} 3 & 1 & 3 \\ 2 & 2 & 3 \\ 2 & 1 & 4 \end{vmatrix}\).

Step 2: Evaluate: \(3(8-3) - 1(8-6) + 3(2-4) = 15 - 2 - 6 = 7\).

Step 3: Write the final answer. \(\boxed{7}\)