# How should I handle the limit with the floor function?

I wanted to show that

$$\lim_{n\to\infty} \bigg( \frac{n+\lfloor x \sqrt{n} \rfloor}{n-\lfloor x \sqrt{n} \rfloor}\bigg)^{\lfloor x \sqrt{n} \rfloor} =e^{2x^2}$$ After applying $x \sqrt{n} -1 \leq \lfloor x \sqrt{n} \rfloor \leq x \sqrt{n}$ and replacing $\sqrt{n}$ with $k$, I tried to squeeze it. It worked out quite nicely with l'Hospital's rule.

Yet soon I realized that given $n\in \mathbb N$ I can neither replace nor use l'Hospital's rule. I recalled Stolz–Cesàro theorem yet it seems to be a nightmare to apply it here.

What should I do in this particular case? Is there a way to justify the replacement and the use of the l'Hospital's rule?

## 2 Answers

After thinking for a while I am ready to answer my own question:

Both the replacement and the use of l'Hospital's rule are justified here since if $\lim_{x\to\infty}f(x)$ exists then $\lim_{n\to\infty}f(n)$ exists as well (since it's merely a subsequence) and is equal the former. Yet the opposite it's true since $\lim_{n\to\infty}\cos(2\pi n)$ exists but $\lim_{x\to\infty}\cos(2\pi x)$ doesn't.

If a limit $\lim_{x\to\infty} f(x) = L$ holds for $x\in \mathbb R$ then also $\lim_{n\to\infty}f(n)=L$ necessarily. Hence you can replace $n$ with $x$ and then $x$ with $k^2$ and IF the continuous result gives a good limit THEN the original limit was the same. The converse does not hold; you can't go from knowing a discrete limit to a continuous one.

The reason for this is that the definition of a real limit $f(y)\to L$ as $y\to x$ implies that for any particular sequence of points $x_n\to x$ tending to that limit we have $f(x_n)\to L$.