$$\lim_{n\rightarrow \infty} \binom{n}{k} h^{(n-k)} , |h| < 1$$
I'm trying to evaluate this limit, but like the $h$ messes me up each time. What I'm doing is trying to prove that the infinite power of a Jordan Matrix has a limit when the eigenvalue is less than one. so in this case, $h$ will be the only Eigenvalue. I have absolute no idea where to go from here.
On
$\lim_{x\rightarrow a}f(x)=f(a)$ and $\lim_{x\rightarrow a}g(x)=g(a)$ where f(a), g(a) are finite, then $\lim_{x\rightarrow a}f(x)g(x)=f(a)g(a)$.
In this problem the for fixed k, the lim of the combinatorics tends to 1 and the lim of the h^(n-k) is 0 then the product of the two is 0.
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Using Stirling's Formula we have
$$\begin{align} \binom{n}{k}&=\frac{n!}{k!(n-k)!}\\\\ &\sim\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{k!\sqrt{2\pi (n-k)}\left(\frac{n-k}{e}\right)^{n-k}}\\\\ &=\frac{n^k\left(1-\frac kn\right)^{k-1/2}}{k!e^k\left(1-\frac kn\right)^{n} }\\\\ &\sim \frac{n^k}{k!} \end{align}$$
as $n\to \infty$. And $\lim_{n\to \infty}\frac{n^k}{k!}h^{n-k}=0$ for $|h|<1$
Assuming that $k$ is fixed, the limit is $0$, since the binomial coefficient increases polynomially in $n$ and $h^n$ decreases exponentially in $n$.