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

lim_(x rarr0^(+))(f(x^(2))-f(sqrt(x)))/(x)

lim_(x rarr0)(6x^(3)-5x^(2)-7x+8)

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 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'(x) be continuous at x=0 and f'(0)=4 then value of lim_(x rarr0)(2f(x)-3f(2x)+f(4x))/(x^(2))

If f'(2)=6 and f'(1)=4 ,then lim_(x rarr0)(f(x^(2)+2x+2)-f(2))/(f(1+x-x^(2))-f(1)) is equal to ?

f(x)=e^x then lim_(x rarr 0) f(f(x))^(1/{f(x)} is