If f(x) = begin{vmatrix} sin x & cos x & tan | vedclass

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(D) We are given $f(x) = \begin{vmatrix} \sin x & \cos x & 	an x \ x^3 & x^2 & x \ 2x & 1 & 1 \end{vmatrix}$.Expanding the determinant along the fi

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(D) We are given $f(x) = \begin{vmatrix} \sin x & \cos x & 	an x \ x^3 & x^2 & x \ 2x & 1 & 1 \end{vmatrix}$.Expanding the determinant along the first row:$f(x) = \sin x(x^2 - x) - \cos x(x^3 - 2x^2) + 	an x(x^3 - 2x^4)$.We need to find $\lim_{x 	o 0} \frac{f(x)}{x^2}$.Dividing each term by $x^2$:$\frac{f(x)}{x^2} = \sin x \left( \frac{x^2 - x}{x^2} ight) - \cos x \left( \frac{x^3 - 2x^2}{x^2} ight) + 	an x \left( \frac{x^3 - 2x^4}{x^2} ight)$.$\frac{f(x)}{x^2} = \sin x \left( 1 - \frac{1}{x} ight) - \cos x (x - 2) + 	an x (x - 2x^2)$.As $x 	o 0$,$\sin x \approx x$ and $	an x \approx x$.$\frac{f(x)}{x^2} \approx x(1 - \frac{1}{x}) - 1(0 - 2) + x(0 - 0) = x - 1 + 2 = x + 1$.Taking the limit as $x 	o 0$,we get $0 + 1 = 1$.

If $f(x) = \begin{vmatrix} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & 1 \end{vmatrix}$,then $\lim_{x \to 0} \frac{f(x)}{x^2}$ is

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