JEE MAIN - Mathematics (2002 - No. 8)
If $$f\left( 1 \right) = 1,{f'}\left( 1 \right) = 2,$$ then
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$$ is
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$$ is
$$2$$
$$4$$
$$1$$
$${1 \over 2}$$
Selitys
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1 } \over {\sqrt x - 1}}\,\,\left( {{0 \over 0}} \right)$$ form using $$L'$$ Hospital's rule
$$ = \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}$$
$$ = {{f'\left( 1 \right)} \over {\sqrt {f\left( 1 \right)} }} = {2 \over 1} = 2.$$
$$ = \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}$$
$$ = {{f'\left( 1 \right)} \over {\sqrt {f\left( 1 \right)} }} = {2 \over 1} = 2.$$