If f(a)=2, f^{prime}(a)=1, g(a)=-1, g^{prime} | vedclass

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(B) We are given the limit: $\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}$.Since $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,substi

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$5$

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(B) We are given the limit: $\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}$.Since $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,substituting $x=a$ gives $\frac{g(a)f(a)-g(a)f(a)}{a-a} = \frac{0}{0}$,which is an indeterminate form.Applying $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:$\lim _{x \rightarrow a} \frac{\frac{d}{dx}[g(x) f(a)-g(a) f(x)]}{\frac{d}{dx}[x-a]}$$= \lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1}$$= g^{\prime}(a) f(a)-g(a) f^{\prime}(a)$$= (2)(2)-(-1)(1)$$= 4+1 = 5$.

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

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