It is formally a question about a limit. Informally, it asks what does p get close to when t gets very big.
[MATH]p = \dfrac{6t^2 + 5t}{(t + 1)^2} = \dfrac{6t^2 + 5t}{t^2 + 2t + 1}.[/MATH]
Suppose t = 10, Then p = (600 + 50)/11^2, about 5.4.
Suppose t = 100. Then p = (60000 + 500)/ 101^2, about 5.9
Suppose t = 1000. Then p = (6,005,000)/1001^2, about 5.99.
So it looks as though the answer is that in the long term, p will approximate 6.
Here is how to confirm that answer in a semi-rigorous way.
[MATH]p = \dfrac{6t^2 + 5t}{t^2 + 2t + 1} = \dfrac{6t^2 + 12t + 6 - 12t - 6 + 5t}{t^2 + 2t + 1} = [/MATH]
[MATH]\dfrac{6(t^2 + 2t + 1)}{t^2 + 2t + 1} + \dfrac{-\ 7t - 6}{t^2 + 2t + 1} = 6 -\dfrac{7t + 6}{(t + 1)^2}.[/MATH]
Now it should be intuitively obvious that the fraction is positive so subtracting it from 6 means p will never reach 6. But it should also be obvious that t squared grows a lot faster than 7t. If t is 1000, then t^2 is a million, which is a lot bigger than 7000. So the fraction is closing in on zero. Thus, the limit is 6 - 0 = 6.
This is not a formal proof, but it gives the idea of what a limit is and how to find it if it exists. (It does not have to exist.)
Hope this helps.