#Can u help me?
void crystalBOT
ripe spruce
1st limit
lim x-> 1 ( f(x) / 1-x )
2nd limit
lim x->2 ( f(x-1) / (x-2) )
here let x-1 = a
then, limit 2 becomes,
lim a->1 ( f(a) / (a-1) )
as u see, 2nd limit is -ve of 1st limit
2nd limit = (-1) 1st limit
hence, 1st limit - 2nd limit = (2) 1st limit = -1
1st limit = -1
also, 3rd limit
lim x->3 ( f(x-2) / (x^2 - 3x) )
taking x common from denoominator
lim x->3 (1/x)( f(x-2) / (x-3) )
taking b= x-2
lim b->1 (1/(b+2) ) ( f(b) / b-1)
here, the 2nd part is same as 1st limit, which is (-1)
hence
lim b->1 (1/(b+2) ) (-1)
which is (-1/3)
jade bison