$P_{k}(n)$ = the number of partitions of n into k parts. Now, if we fix some $t\ge 0$ , then $\lim_{n\to\infty}P_{n-t}(n)\to$ c, c being some constant.
Please help me with this! I believe $P_{k}(n)=\dbinom{n+k-1}{n}$. If that be so then, $P_{n-t}(n)$ comes out to some horrible combinatorial and I can't proceed any further to show that it'll be bounded by a constant. Also I'll be extremely glad if you could provide me more insight into the result.
First let me observe that the OP has the formula for compositions as opposed to partitions. Assuming the OP refers to partitions there is no need for calculus or Stirling's approximation. As we are considering the limit as $n$ goes to infinity we may assume that $n\ge 2t.$ Imagine the partition as a sequence of adjacent columns of stacked boxes, with the height of every stack being the value of the part. As we have $n-t$ parts $\ge 1$ we get a minimal contribution of $n-t$ from those parts and this is the bottom row. That leaves $t$ boxes to be distributed into the $n-t$ columns and since $n-t\ge t$ there are $P(t)$ configurations (themselves partitions) we can do so taking symmetry into account. That is our constant, it has the value $P(t).$ QED.