Is it the case that $\frac{k_n}{\sqrt{n}}\to 0\Longrightarrow\left(1-\frac{k_n}{n}\right)^{k_n}\to 1$? Suppose that $(k_n)_{n\in\mathbb N}$ is a sequence of positive real numbers such that $k_n/\sqrt{n}\to 0$ as $n\to\infty $. I want to show that in this case $$\lim_{n\to\infty}\left(1-\frac{k_n}{n}\right)^{k_n}=1.$$

I made some experimentation with sequences of the form $k_n\equiv n^{\alpha}$ for $\alpha\in(0,1)$. It seems to be the case that the limit above is $0$ if $\alpha<1/2$ (which is a special case of the statement I want to prove), and the limit is $1$ if $\alpha>1/2$. The knife-edge case is easy: $\alpha=1/2$ implies that the limit is $1/\mathbb e $. 

However, I am not sure how to proceed to prove the general case with of $k_n/\sqrt{n}\to0$. Any hints would be appreciated.
 A: As $x\to 0$ we have $\log(1-x) = - x + \mathcal{O}(x^2)$ so as long as $\frac{k_n}{n} \to 0$ it follows that
$$\left(1 - \frac{k_n}{n}\right)^{k_n} = e^{k_n\log\left(1 - \frac{k_n}{n}\right)} = e^{-\left(\frac{k_n}{\sqrt{n}}\right)^2 + \mathcal{O}\left(\frac{k_n^3}{n^2}\right)}$$
The limit you are after is therefore
$$\lim_{n\to\infty}\left(1 - \frac{k_n}{n}\right)^{k_n} = \lim_{n\to\infty}e^{-\left(\frac{k_n}{\sqrt{n}}\right)^2} = e^{-\lim_{n\to\infty} \left(\frac{k_n}{\sqrt{n}}\right)^2}$$
From this general form it follows that if $k_n = n^\alpha$ with $\alpha\in(0,1)$ then the limit is $0$ if $\alpha > \frac{1}{2}$, $e^{-1}$ if $\alpha = \frac{1}{2}$ and $1$ if $\alpha < \frac{1}{2}$.
A: The expression equals
$$\tag 1\left[\left (1-\frac{k_n}{n}\right )^{n/k_n}\right ]^{k_n^2/n}.$$
Now $k_n/\sqrt n \to 0$ implies $(k_n/\sqrt n)^2 = k_n^2/n \to 0$ as well as $k_n/n \to 0.$ Because $(1-u)^{1/u}\to 1/e$ as $u\to 0,$ the limit inside the brackets in $(1)$ is $1/e.$ The limit for our expression is therefore $(1/e)^0 = 1.$
A: I think I got it. Fix any $\varepsilon\in(0,1)$. Since $$\lim_{n\to\infty}\frac{k_n}{\sqrt{n}}=0,$$ there exists some $N\in\mathbb N$ such that for all positive integers $n$ exceeding $N$, one has that $k_n<\varepsilon\sqrt{n}<\sqrt{n}\leq n$. Clearly, $$\left(1-\frac{k_n}{n}\right)^{k_n}\leq 1$$ holds for all such $n\in\mathbb N$, so that $$\limsup_{n\to\infty}\left(1-\frac{k_n}{n}\right)^{k_n}\leq 1.$$
On the other hand, if $n\in\mathbb N$ and $n> N$, then $$\left(1-\frac{k_n}{n}\right)^{k_n}\geq\left(1-\frac{\sqrt{n}}{n}\right)^{k_n}=\left(1-\frac{1}{\sqrt{n}}\right)^{k_n}\geq\left(1-\frac{1}{\sqrt{n}}\right)^{\varepsilon \sqrt{n}},$$ where the last inequality follows because the base satisfies $1-1/\sqrt{n}< 1$. As a consequence, $$\liminf_{n\to\infty}\left(1-\frac{k_n}{n}\right)^{k_n}\geq\liminf_{n\to\infty}\left(1-\frac{1}{\sqrt{n}}\right)^{\varepsilon \sqrt{n}}=\left(\frac{1}{\mathbb e}\right)^{\varepsilon}.$$ As $\varepsilon$ can be made arbitrarily close to $0$, the claim follows.
