" If "f(2)=4" and "f(2)=1" then "underset x rarr2Lt(x)/(x)(x(2)-2f(x))/(x-2)

if f(2)=4 and f'(2)=2 then prove that underset(xrarr2)Lt (2x^2-2f(x))/(x-2)=4

If f(2)=4 and f'(2)=1, then find (lim)_(x rarr2)(xf(2)-2f(x))/(x-2)

If f(2)=4,f'(2)=4, then value of underset(x rarr 2)lim(xf(2)-2f(x))/(x-2) is

If f(2)=4 ,f'(2)=1 , underset(xrarr2)Lt ((xf(2)-2f(x))/(x-2)) =?

If f(2)=2 and f'(2)=1, and then underset(x to 2) lim (2x^(2)-4f(x))/(x-2) is equal to

If f(2) = 4 and f^(')(2) =1 then lim_(x rarr 2) (x f(2)-2f (x))/(x-2)

If f(2) = 2 and f^(')(2) = 1 , then lim_(x rarr 2) (2x^(2)-4 f(x))/(x-2) =

If f is a real function such that f(4) = 4 and f'(4) = 16, " then " underset(x rarr 4)(lim) (sqrt(f(x))-2)/(sqrtx-2)=

If f(2)=4,f'(2)=4 , then value of underset(x to 2)lim(xf(2)-2f(x))/(2(x-2)) is -