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 =

lim_(x rarr0)x^(2)+3x+3

lim_(x rarr0)(2^(x)-3^(x))/(x)

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

If f'(x)=k,k!=0, then the value of lim_(x rarr0)(2f(x)-3f(2x)+f(4x))/(x^(2)) is

lim_(x rarr 0) x^2/y = 0

lim_(x rarr0)(sqrt(x+4)-2)/(x)

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

If the normal of y=f(x) at (0,0) is given by y-x=0, then lim_(x rarr0)(x^(2))/(f(x^(2))-20f(9x^(2))+2f(99x^(2))) is equal