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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Aicharya · Accepted Answer

Applying L-Hospital's rule, we get $\begin{aligned} & \lim _{x ightarrow a} \frac{g(x) f(a)-g(a) f(x)}{(x-a)} \ & =\lim _{x ightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1} \ & =g^{\prime}(a) f(a)-g(a) f^{\prime}(a) \ & =2(2)-(-1)(1) \ & =4+1 \ & =5 \end{aligned}$

If $\mathrm{f}(\mathrm{a})=2, \mathrm{f}^{\prime}(\mathrm{a})=1, \mathrm{~g}(\mathrm{a})=-1,…

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