If f(1) = 1 and f'(1) = 2,then mathop {lim }l | vedclass

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(A) Let $y = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Multiplying the numerator and denominator by the conjugates $(

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$2$

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(A) Let $y = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Multiplying the numerator and denominator by the conjugates $(\sqrt{f(x)} + 1)$ and $(\sqrt{x} + 1)$:$y = \mathop {\lim }\limits_{x 	o 1} \frac{(\sqrt{f(x)} - 1)(\sqrt{f(x)} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} 	imes \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$$y = \mathop {\lim }\limits_{x 	o 1} \frac{f(x) - 1}{x - 1} 	imes \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$Since $f(1) = 1$,we can write $f(x) - 1$ as $f(x) - f(1)$:$y = \left( \mathop {\lim }\limits_{x 	o 1} \frac{f(x) - f(1)}{x - 1} ight) 	imes \left( \mathop {\lim }\limits_{x 	o 1} \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1} ight)$$y = f'(1) 	imes \frac{\sqrt{1} + 1}{\sqrt{f(1)} + 1} = 2 	imes \frac{2}{1 + 1} = 2 	imes 1 = 2$.Alternatively,using $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{2\sqrt{f(x)}} f'(x)}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x 	o 1} \frac{f'(x) \sqrt{x}}{\sqrt{f(x)}} = \frac{f'(1) \sqrt{1}}{\sqrt{f(1)}} = \frac{2 	imes 1}{1} = 2$.

If $f(1) = 1$ and $f'(1) = 2$,then $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

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