Show that $\lim\limits_{x \to \infty} \frac{\lfloor x \rfloor}{x} = 1$

Question

I know that $x-1 < \lfloor x \rfloor \leq x$ hence $1-\dfrac{1}{x}<\dfrac{\lfloor x \rfloor}{x} \leq 1$ for $x>0$. I think it is easy to see from this that the limit should be 1 but I don't know how to formally prove this.

Answer 1

You know that $1-\dfrac{1}{x}<\dfrac{\lfloor x \rfloor}{x} \leq 1$ for any $x > 0$ so taking the limits of the leftmost expression and of the rightmost expression you get:

$\lim_{x \to \infty} 1-\dfrac{1}{x} = 1$

and

$\lim_{x \to \infty} 1 = 1$

And since $\dfrac{\lfloor x \rfloor}{x}$ is always in between, by the squeeze theorem you get

$$\lim_{x \to \infty} 1-\dfrac{1}{x} \leq \lim_{x \to \infty} \dfrac{\lfloor x \rfloor}{x} \leq \lim_{x \to \infty} 1 \Rightarrow$$

$$\lim_{x \to \infty} \dfrac{\lfloor x \rfloor}{x} = 1$$