Show that :
(xax−b)a−b.(xbx−c)b−c.(xcx−a)c−1=1
L.H.S.(xax−b)a−b.(xbx−c)b−c.(xcx−a)c−1=(xa+b)a−b.(xb+c)b−c.(xc+a)c−a=x(a+b)(a−b).x(b+c)(b−c).x(c+a)(c−a)=xa2−b2+b2−c2+c2−a2=x0=1=R.H.S.
Simplify :
(xa+b)a−b.(xb+c)b−c.(xc+1)c−a
Show that limx→∞(√x2+x+1−x)≠limx→∞(√x2+1−x)