Let  a1>a2>a3>...>an>1.p1>p2>p3>....>pn>0   such  that   p1+p2+p3+.....pn =1.Also,   Fx=p1a1x+p2a2x+....+pnanx1xlimx→0+F(x)   equalslimx→∞F(x)   equals tolimx→-∞F(x)   equals to

limx→0+  Fx=  limx→0+p1a1x+p2a2x+....+pnanx1x      1∞  form               =  elimx→0+   p1a1x+p2a2x+....+pnanx-1x               =  elimx→0+   p1a1x  ln  a1+ p2a2x   ln  a2+.....+pnanxln   an                =  ep1  ln  a1+p2   ln   a2+....+p2   ln   an               =   eln   a1p1a2p2.....anpn=a1p1.a2p2a3p3....anpnlimx→∞  Fx=L=limx→∞  p1a1x+p2a2x+....pnanx1x         ∞0  form∴   log  L=limx→∞  logep1a1x+p2a2x+....pnanxxUsing   L'Hospital's  rule log  L= limx→∞  p1a1x  ln  a1+p2a2x ln  a2+....+  pnanx   ln  anp1a1x+p2a2x+...+pnpnx    1Dividing  by   a1x   and   taking   limit,  we   get limx→∞   a2a1x=0,  limx→∞a3a1x=0  etc... log  L=p1  ln  a1p2=ln  a1or   L=a1Let  limx→-∞  Fx=L∴   log  L=  limx→-∞  p1a1x  lna1+p2a2x  ln  a2+....  +  pnanx  ln  anp1a1x+p2a2x+...+pnanxDividing   by   anx   and   taking  limx→-∞   a1anx,a2anx,   etc.   vanish.Therefore,        log  L=  pn ln  anpnor       L=an