If f(4) = g(4) = 2,f'(4) = 9,and g'(4) = 6,th | vedclass

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(A) Given the limit $L = \mathop {	ext{Limit}}\limits_{x 	o 4} \frac{\sqrt{f(x)} - \sqrt{g(x)}}{\sqrt{x} - 2}$.Since the form is $\frac{\sqrt{2} - \

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$3\sqrt{2}$

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(A) Given the limit $L = \mathop {	ext{Limit}}\limits_{x 	o 4} \frac{\sqrt{f(x)} - \sqrt{g(x)}}{\sqrt{x} - 2}$.Since the form is $\frac{\sqrt{2} - \sqrt{2}}{\sqrt{4} - 2} = \frac{0}{0}$,we apply $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$L = \mathop {	ext{Limit}}\limits_{x 	o 4} \frac{\frac{d}{dx}(\sqrt{f(x)} - \sqrt{g(x)})}{\frac{d}{dx}(\sqrt{x} - 2)}$$L = \mathop {	ext{Limit}}\limits_{x 	o 4} \frac{\frac{f'(x)}{2\sqrt{f(x)}} - \frac{g'(x)}{2\sqrt{g(x)}}}{\frac{1}{2\sqrt{x}}}$Substitute $x = 4$:$L = \frac{\frac{f'(4)}{2\sqrt{f(4)}} - \frac{g'(4)}{2\sqrt{g(4)}}}{\frac{1}{2\sqrt{4}}}$$L = \frac{\frac{9}{2\sqrt{2}} - \frac{6}{2\sqrt{2}}}{\frac{1}{4}}$$L = \frac{\frac{3}{2\sqrt{2}}}{\frac{1}{4}} = \frac{3}{2\sqrt{2}} 	imes 4 = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

If $f(4) = g(4) = 2$,$f'(4) = 9$,and $g'(4) = 6$,then $\mathop {\text{Limit}}\limits_{x \to 4} \frac{\sqrt{f(x)} - \sqrt{g(x)}}{\sqrt{x} - 2}$ is equal to:

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