Let $f(x) = \begin{vmatrix} x \cos x & 2x \sin x & x \tan x \\ 1 & 2x & 1 \end{vmatrix}$, then $\lim_{x \to 0} \frac{f(x)}{x^2} =$
Step-by-Step Solution
Key Concept: Factor out x from the first row of the determinant to get f(x) = x·g(x), then apply L'Hôpital's rule or Taylor series expansions (cos x ≈ 1 - x²/2, sin x ≈ x, tan x ≈ x) to evaluate the limit of g(x)/x as x → 0.
Given $f(x) = \begin{vmatrix} x\cos x & 2x\sin x & x\tan x \\ 1 & x & 1 \\ 1 & 2x & 1 \end{vmatrix}$, rewrite as $f(x) = \begin{vmatrix} \cos x & 2\sin x & \tan x \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix}$. Evaluate $\lim_{x \to 0} \frac{f(x)}{x^2}$ using $R_1 - R_2$ to get $\begin{vmatrix} 1 & 2 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} = -1$.
Correct Answer: 3