If f(a) = 2,f'(a) = 1,g(a) = -3,g'(a) = -1,th | vedclass

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(A) Given the limit $\mathop {\lim }\limits_{x 	o a} \frac{f(a)g(x) - f(x)g(a)}{x - a}$.Substituting $x = a$,we get $\frac{f(a)g(a) - f(a)g(a)}{a - a

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$1$

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(A) Given the limit $\mathop {\lim }\limits_{x 	o a} \frac{f(a)g(x) - f(x)g(a)}{x - a}$.Substituting $x = a$,we get $\frac{f(a)g(a) - f(a)g(a)}{a - a} = \frac{0}{0}$ form.Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o a} \frac{\frac{d}{dx}(f(a)g(x) - f(x)g(a))}{\frac{d}{dx}(x - a)} = \mathop {\lim }\limits_{x 	o a} \frac{f(a)g'(x) - f'(x)g(a)}{1}$.Now,substitute $x = a$:$= f(a)g'(a) - f'(a)g(a)$.Given $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$:$= (2)(-1) - (1)(-3) = -2 + 3 = 1$.

If $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$,then $\mathop {\lim }\limits_{x \to a} \,\frac{f(a)g(x) - f(x)g(a)}{x - a} = $

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