If `f(0)=3` then `Lt_(x->oo) x^2/(f(x^2)-6,f(4x^2)+5.f(7x^2)`

If f^(1)(0)=-3 then Lt_(x to 0) (x^(2))/(f(x^(2))-6.f(4x^(2))+5.f(7x^(2)))=

If f'(0)=3 then lim_(x rarr0)(x^(2))/(f(x^(2))-6f(4x^(2))+5f(7x^(2))=)

If the normal to differentiable curve y = f(x) at x = 0 be given by the equation 3x-y + 3 = 0 , then the value of lim_(x->0) x^2(f(x^2)-5f(4x^2)+ 4f(7x^2)) is

If the equation of the normal to the curve y=f(x) at x=0 is 3x-y+3=0 then the value of lim_(xrarr0) (x^(2))/({f(x^(2))-5f(4x^(2))+4f(7x^(2))}) is

If the equation of the normal to the curve y=f(x) at x =0 is 3x-y+3=0 then the value of lim_(x rarr0)(x^(2))/({f(x^2)-5f(4x^(2))+4f(7x^(2))}) is

If the normal to differentiable curve y=f(x) at x=0 be given by the equation 3x-y+3=0, then the value of lim_(x rarr0)(x^(2))/(f(x^(2))-5f(4x^(2))+4f(7x^(2))) is

If the normal to the curve y=f(x) at x=0 be given by the equation 3x-y+3=0 , then the value of lim_(x rarr0)x^(2){f(x^(2))-5f(4x^(2))+4f(7x^(2))}^(-1) is (-1)/(k) then k =

Let f: R to R be a positive increasing function with Lt_(x to oo)(f(3x))/(f(x))=1 then Lt_(x to oo)(f(2x))/(f(x))=