Let f(x) = 3-x^2 when -1<=x<2
And 2x-4 when 2<=x<=4
Now in book when they are calculating left hand derivative they do like that
[f (2-h)-f(2)]/h h----0
Here they have not taken f (2) =0(reason being 2 is not in domain of left hand function) so they take f (2) as limit value =-1
Hence they calculate LHD =-4
sir my doubt is that when we define tangent ,we take two point P and Q. in which one point is always fixed (P) and other always try to approach(Q) .so we should take exact value of f(2)=0 ,as f is defined at 2.
ONE MORE IMPORTANT POINT
derivative of f(x) at point c is defined as [ f(x)-f(c)]/(x-c) where x------c.
here f(c) is value of f at exact value of c or limiting value?