If f(x) = left| begin{array}{ccc} 1 & 6+x & 3 | vedclass

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(C) Given $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \ 0 & x-3 & 3x^2-27 \ 0 & 2x-4 & 8x^2-32 \end{array} ight|$.Expanding along the first

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(C) Given $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \ 0 & x-3 & 3x^2-27 \ 0 & 2x-4 & 8x^2-32 \end{array} ight|$.Expanding along the first column:$f(x) = 1 \cdot [(x-3)(8x^2-32) - (2x-4)(3x^2-27)]$$f(x) = (8x^3 - 32x - 24x^2 + 96) - (6x^3 - 54x - 12x^2 + 108)$$f(x) = 2x^3 - 12x^2 + 22x - 12 = 2(x-1)(x-2)(x-3)$.Now,$f(-x) = 2(-x-1)(-x-2)(-x-3) = -2(x+1)(x+2)(x+3)$.Evaluating the limit:$\lim_{x ightarrow 1} \frac{f(x)}{f(-x)} = \lim_{x ightarrow 1} \frac{2(x-1)(x-2)(x-3)}{-2(x+1)(x+2)(x+3)}$.Substituting $x=1$:$= \frac{2(1-1)(1-2)(1-3)}{-2(1+1)(1+2)(1+3)} = \frac{0}{-48} = 0$.

If $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$,then $\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = $

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