How to find $\lim_{x \to \infty} [x]/x$?

Question

Find the limit $$\lim_{x\to \infty } \ \frac {[x]}{x}.$$

Does $[x]$ means greatest integer in this case?

Answer 1

first we have $\ \frac {[x]}{x} = \ \frac {(x-1)+k}{x}$ for $k \in [0,1)$

$$\lim_{x\to \infty } \ \frac {[x]}{x}= \lim_{x\to \infty } \frac {x-1}{x}+ \lim_{x\to \infty }\frac {k}{x}=1$$

Answer 2

$$\lim_{x \to \infty} \left( \frac{x}{x} - \frac{\lfloor{x}\rfloor}{x} \right) = \lim_{x \to \infty} \frac{x - \lfloor{x}\rfloor}{x} $$

But $0 \leq x - \lfloor{x}\rfloor < 1$, as $x - \lfloor{x}\rfloor$ is the fractional part of $x$ so:

$$0 \leq \lim_{x \to \infty} \frac{x - \lfloor{x}\rfloor}{x} \leq \lim_{x \to \infty} \frac{1}{x} = 0$$

Then this implies that (by the Squeeze Theorem):

$$\lim_{x \to \infty} \frac{x - \lfloor{x}\rfloor}{x} = 0$$

$$\therefore \lim_{x \to \infty} \left( \frac{x}{x} - \frac{\lfloor{x}\rfloor}{x} \right) = 0$$

But we know:

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

Hence, for the solution to hold:

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

Answer 3

Hint:

$$\frac{x-1}{x}\leq \frac{\lfloor x \rfloor}{x} \leq \frac{x}{x}$$

Now squeeze.

Answer 4

We know that $\left|x-[x]\right|<1$ always. Now , $$\left|\frac{[x]}{x}-1\right|=\left|\frac{[x]-x}{x}\right|<\frac{1}{x}<\epsilon \text{ , whenever } x>\frac{1}{\epsilon}.$$Taking $\displaystyle N=\left[\frac{1}{\epsilon}\right]+1$ we get, $\displaystyle \left|\frac{[x]}{x}-1\right|<\epsilon$ whenever , $x\ge N$.