If f(x) = frac{2}{x - 3},g(x) = frac{x - 3}{x | vedclass

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(C) Given $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$,and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$.Note that $x^2 + x - 12 = (x - 3)(x + 4)

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$ - \frac{2}{7}$

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(C) Given $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$,and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$.Note that $x^2 + x - 12 = (x - 3)(x + 4)$.Summing the functions: $f(x) + g(x) + h(x) = \frac{2}{x - 3} + \frac{x - 3}{x + 4} - \frac{2(2x + 1)}{(x - 3)(x + 4)}$.Taking the common denominator $(x - 3)(x + 4)$:$= \frac{2(x + 4) + (x - 3)^2 - 2(2x + 1)}{(x - 3)(x + 4)}$$= \frac{2x + 8 + x^2 - 6x + 9 - 4x - 2}{(x - 3)(x + 4)}$$= \frac{x^2 - 8x + 15}{(x - 3)(x + 4)}$$= \frac{(x - 3)(x - 5)}{(x - 3)(x + 4)}$$= \frac{x - 5}{x + 4}$ for $x 
eq 3$.Therefore,$\lim_{x 	o 3} [f(x) + g(x) + h(x)] = \lim_{x 	o 3} \frac{x - 5}{x + 4} = \frac{3 - 5}{3 + 4} = - \frac{2}{7}$.

If $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$ and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$,then $\lim_{x \to 3} [f(x) + g(x) + h(x)]$ is

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