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

Options

A. `-3`
B. `1//3`
C. 3
D. `-1//3`

Answer

D

Explanation

`y=3x+3` is normal to the curve at x = 0
Slope of normal is 3.
`therefore" Slope of tangent at x = 0 is "f'(0)=-(1)/(3)`
Now `underset(xrarr0)(lim)(x^(2))/({f(x^(2))-5f(4x^(2))+4f(7x^(2))})((0)/(0)" form")`
`=underset(xrarr0)(lim)(2x)/(2xf'(x^(2))-40xf'(4x^(2))+56xf'(7x^(2)))`
`=underset(xrarr0)(lim)(1)/(f'(0)-20f'(0)+28f'(0))`
`=(1)/(9f'(0))=(1)/(9(-(1)/(3)))=-(1)/(3)`