The correct option is D limx→−∞F(x)=an
limx→0F(x)=limx→0(p1ax1+p2ax2+⋯+pnaxn)1/x
=elimx→0⎛⎜⎝p1ax1+p2ax2+⋯+pnaxn−1x⎞⎟⎠
=elimx→0(p1ax1lna1+p2ax2lna2+⋯+pnaxnlnan)
=e(p1lna1+p2lna2+⋯+pnlnan)
=e(lnap11+lnap22+⋯+lnapnn)
=e(lnap11ap22⋯apnn)
=ap11ap22⋯apnn
L=limx→∞F(x)=limx→∞(p1ax1+p2ax2+⋯+pnaxn)1/x (∞0 form)
⇒lnL=limx→∞ln(p1ax1+p2ax2+⋯+pnaxn)x
Using L'Hopitals' rule, we get
lnL=limx→∞p1ax1lna1+p2ax2lna2+⋯+pnaxnlnanp1ax1+p2ax2+⋯+pnaxn
Multiplying and dividing by ax1 in RHS, we get
(a2a1)x,(a3a1)x, etc. all vanish as x→∞ because a1>a2>a3>…an>1
⇒lnL=p1lna1p1=lna1
Hence, lnL=lna1
⇒L=a1
Let limx→−∞F(x)=L1
∴lnL1=limx→∞p1ax1lna1+p2ax2lna2+⋯+pnaxnlnanp1ax1+p2ax2+⋯+pnaxn
Multiplying and dividing by (an)x in RHS (a1an)x,(a2an)x, etc. vanish as x→−∞
⇒lnL1=pnlnanpn
⇒L1=an