If f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2, | vedclass

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(C) Let $L = \lim_{x 	o a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$.Since the limit is in the form $\frac{0}{0}$ as $x 	o a$,we apply $L$'Hospital's rule

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$5$

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(C) Let $L = \lim_{x 	o a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$.Since the limit is in the form $\frac{0}{0}$ as $x 	o a$,we apply $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:$L = \lim_{x 	o a} \frac{\frac{d}{dx}[g(x)f(a) - g(a)f(x)]}{\frac{d}{dx}[x - a]}$$L = \lim_{x 	o a} \frac{g'(x)f(a) - g(a)f'(x)}{1}$Now,substitute $x = a$:$L = g'(a)f(a) - g(a)f'(a)$Given $f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2$:$L = (2)(2) - (-1)(1)$$L = 4 + 1 = 5$.

If $f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2$,then the value of $\lim_{x \to a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$ is:

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