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?

Solution 1:

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

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

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

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:

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