If f(1) = 1 and f'(1) = 4, then the value of | vedclass

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(B) Given the limit $L = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôp

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$4$

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(B) Given the limit $L = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:$L = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(\sqrt{f(x)} - 1)}{\frac{d}{dx}(\sqrt{x} - 1)}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$Substituting $x = 1$:$L = \frac{f'(1) \cdot \sqrt{1}}{\sqrt{f(1)}} = \frac{4 \cdot 1}{1} = 4$.

If $f(1) = 1$ and $f'(1) = 4,$ then the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

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