I have a trouble finding asymptotic for $${n}\choose{k}$$ if $$k^{2} = o(n)$$
I've tried and got $$\frac{n(n-1)...(n-k+1)}{k!},$$ then I tried to use Stirling's formula or somehow use that we have $$\lim_{n \to \infty} \frac{k^{2}}{n} = 0,$$ but I can't understand how to use it, pls help me out
Let's take \begin{align*} \binom{n}{k} & = \frac{n!}{k!(n-k)!} = \frac{n(n-1)\ldots (n-k+1)}{k!} = \frac{n^k}{k!}\cdot \frac{n}{n}\cdot \frac{n-1}{n}\cdot \ldots \cdot \frac{n-k+1}{n}=\frac{n^k}{k!}\cdot \prod_{j=0}^{k-1} \frac{n-j}{n}. \end{align*}
We will show that, with $k^2=o(n)$, $\lim_n\prod_{j=0}^{k-1}\frac{n-j}{n}=1$. This would be easy for constant $k$, since each factor is going to $1$. But if $k$ grows, we could have lots of factors whose product is still far away from $1$.
We note that $0< \frac{n-j}{n}\leqslant 1$ for all $0\leqslant k\leqslant n$ and $0\leqslant j<k$. So $$\prod_{j=0}^{k-1}\frac{n-j}{n}\leqslant 1.$$ We note that $$1\geqslant \prod_{j=0}^{k-1}\frac{n-j}{n} > \prod_{j=0}^{k-1}\Bigl(\frac{n-k}{n}\Bigr)=\Bigl(1-\frac{k}{n}\Bigr)^k.$$ So by the squeeze theorem, it is sufficient to know that $\lim_n (1-k/n)^k=1$ for $k^2=o(n)$.
We note that, usually by definition, $$\lim_x \Bigl(1+\frac{1}{x}\Bigr)^x=e.$$ From this we see that, using the substitution $x=y+1$, \begin{align*} \lim_x \Bigl(1-\frac{1}{x}\Bigr)^x & =\lim_y \Bigl(1-\frac{1}{y+1}\Bigr)^{y+1} = \lim_y \Bigl(1-\frac{1}{y+1}\Bigr)^y \cdot \Bigl(1-\frac{1}{y+1}\Bigr) \\ & = \lim_y \Bigl(1-\frac{1}{y+1}\Bigr)^y \cdot 1 = \lim_y \Bigl(1-\frac{1}{y+1}\Bigr)^y \\ & = \lim_y \Bigl(\frac{y}{y+1}\Bigr)^y = \Bigl[\lim_y \Bigl(\frac{y+1}{y}\Bigr)^y\Bigr]^{-1} = e^{-1}. \end{align*} Of course, $\lim_x \Bigl(1+\frac{0}{x}\Bigr)^x=1$. Therefore for any $r\in\mathbb{R}$, $\lim_x \Bigl(1+\frac{r}{x}\Bigr)^x=e^r$. We obtain this trivially if $r=0$ and otherwise we let $y=x/|r|$ and choose either the $+$ or $-$ case above and substitute.
Therefore if $k=o(n)$, we have \begin{align*} 1 \geqslant \underset{n}{\lim\inf} \Bigl(1-\frac{k}{n}\Bigr)^k \geqslant \sup_{r>0} \underset{n}{\lim\inf} \Bigl(1-\frac{r}{n}\Bigr)^n = e^{-r}.\end{align*} Therefore $$\Bigl(1-\frac{k}{n}\Bigr)^k=1$$ if $k^2=o(n)$.