@titun,
You were right till
For x = 0 the above limit is in 0/0 form. So just apply L. Hospital
Then we will get lnP=
[x ]
[0 ]2*3*(a
xlna+b
xlnb+c
xlnc)/(a
x+b
x+c
x)
Hence P=(abc)2/3
You can also solve this by making use of
[x ][0 ](1+x)
1/x=e but using L'Hospitals rule as done by titun is the shortest method
Moderation Team
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