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

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

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

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

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