If f(9) = 9 and f'(9) = 4,then mathop {lim }l | vedclass

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(B) Given the limit $\mathop {\lim }\limits_{x 	o 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.Since $f(9) = 9$,the expression takes the indeterminat

Vedclass · Accepted Answer

$4$

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(B) Given the limit $\mathop {\lim }\limits_{x 	o 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.Since $f(9) = 9$,the expression takes the indeterminate form $\frac{0}{0}$ as $x 	o 9$.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 9} \frac{\frac{d}{dx}(\sqrt{f(x)} - 3)}{\frac{d}{dx}(\sqrt{x} - 3)} = \mathop {\lim }\limits_{x 	o 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x 	o 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$.Substituting the values $f(9) = 9$ and $f'(9) = 4$:$= \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

If $f(9) = 9$ and $f'(9) = 4$,then $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}} = $

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