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

f(x) and g(x) are two differentiable functions on [0, 2] such that

f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9

then f(x) - g(x) at x = $${3 \over 2}$$ is

Let $$f(2) = 4$$ and $$f'(x) = 4.$$

Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by