Question 289 of 401MediumMCQ
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
Solutions · 1
Official solution
\[\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1 } \over {\sqrt x - 1}}\,\,\left( {{0 \over 0}} \right)\] form using L′ Hospital's rule
\[ = \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}\]
\[ = {{f'\left( 1 \right)} \over {\sqrt {f\left( 1 \right)} }} = {2 \over 1} = 2.\] Answer the question to unlock