Question

Let \[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\] \[\lim_{x \to 0} \frac{f\left( x \right)}{x^2}\]  is equal to

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Answer 1

\[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\]
\[ = \begin{vmatrix}\cos x & x & 1 \\ \sin x & 0 & x \\ \sin x & x & x\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\]
\[ = \begin{vmatrix}\cos x & x & 1 \\ \sin x & 0 & x \\ \sin x - \cos x & 0 & x - 1\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\]
\[ = - x\left[ x \sin x - \sin x - x \sin x + x \cos x \right]\]
\[ = - x\left( x \cos x - \sin x \right)\]
\[ \therefore \lim_{x \to 0} \frac{f\left( x \right)}{x^2} = \lim_{x \to 0} \frac{x\left( \sin x - x \cos x \right)}{x^2}\]
\[ = \lim_{x \to 0} \frac{\sin x}{x} - \lim_{x \to 0} \cos x\]
\[ = 1 - 1 = 0\]