How to find $\lim_{x \to \infty} [x]/x$? Find the limit 
$$\lim_{x\to \infty } \ \frac {[x]}{x}.$$
Does $[x]$ means greatest integer in this case?
 A: first we have $\ \frac {[x]}{x} = \ \frac {(x-1)+k}{x}$ for $k \in [0,1)$
$$\lim_{x\to \infty } \ \frac {[x]}{x}= \lim_{x\to \infty }  \frac {x-1}{x}+ \lim_{x\to \infty }\frac {k}{x}=1$$
A: $$\lim_{x \to \infty} \left( \frac{x}{x} - \frac{\lfloor{x}\rfloor}{x}  \right) = \lim_{x \to \infty} \frac{x - \lfloor{x}\rfloor}{x} $$
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:
$$\lim_{x \to \infty} \frac{\lfloor{x}\rfloor}{x} = 1$$
A: 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 , $x\ge N$.
A: Hint: 
$$\frac{x-1}{x}\leq \frac{\lfloor x \rfloor}{x} \leq \frac{x}{x}$$
Now squeeze.
