Let f(x) = left| begin{array}{ccc} cos x & x | vedclass

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(B) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \ 2 \sin x & x^3 & 2x \ 	an x & x & 1 \end{array} ight|$.Expanding the determinant alo

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$-2$

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(B) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \ 2 \sin x & x^3 & 2x \ 	an x & x & 1 \end{array} ight|$.Expanding the determinant along the first row:$f(x) = \cos x(x^3 - 2x^2) - x(2 \sin x - 2x 	an x) + 1(2x \sin x - x^3 	an x)$.$f(x) = (x^3 - 2x^2) \cos x - 2x \sin x + 2x^2 	an x + 2x \sin x - x^3 	an x$.$f(x) = (x^3 - 2x^2) \cos x + 2x^2 	an x - x^3 	an x$.Now, we need to find $\lim_{x ightarrow 0} \frac{f(x)}{x^2} = \lim_{x ightarrow 0} \frac{(x^3 - 2x^2) \cos x + 2x^2 	an x - x^3 	an x}{x^2}$.$= \lim_{x ightarrow 0} \left( (x - 2) \cos x + 2 	an x - x 	an x ight)$.Substituting $x = 0$:$= (0 - 2) \cos(0) + 2 	an(0) - 0 \cdot 	an(0) = -2(1) + 0 - 0 = -2$.

Let $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{array} \right|$. Then, find the value of $\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$.

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