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If $\mathrm{f}(\mathrm{a})=2, \mathrm{f}^{\prime}(\mathrm{a})=1, \mathrm{~g}(\mathrm{a})=-1, \mathrm{~g}^{\prime}(\mathrm{a})=2$, then as $x$ approaches a, $\frac{\mathrm{g}(x) \mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a}) \mathrm{f}(x)}{(x-\mathrm{a})}$ approaches
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The correct answer is:
5
Applying L-Hospital's rule, we get
$\begin{aligned}
& \lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{(x-a)} \\
& =\lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1} \\
& =g^{\prime}(a) f(a)-g(a) f^{\prime}(a) \\
& =2(2)-(-1)(1) \\
& =4+1 \\
& =5
\end{aligned}$
$\begin{aligned}
& \lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{(x-a)} \\
& =\lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1} \\
& =g^{\prime}(a) f(a)-g(a) f^{\prime}(a) \\
& =2(2)-(-1)(1) \\
& =4+1 \\
& =5
\end{aligned}$
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