limit of function: $ \lim_{x \to \infty} \frac{[x]}{x}$ This is my homework:

Limit of functions: ($[x]$ is the total part of $x$)
  
  
*
  
*$\displaystyle \lim_{x \to \infty} \frac{[x]}{x}$
  
*$\displaystyle \lim_{x \to -\infty} \frac{[x]}{x}$
  
*$\displaystyle \lim_{x \to 0} \frac{[x]}{x}$

For the first one, I used the inequality $$x -1 \leq [x] \leq x.$$
So, $$\frac{1-\frac{1}{x}}{1}=\frac{x-1}{x} \leq \frac{[x]}{x} \leq \frac{x}{x}=1.$$
So my answer is $\displaystyle \lim _{x \to \infty} \frac{[x]}{x} = 1$.
Is that good?
I don't know what to use in other examples$\ldots$
 A: One way I think about this problem is using a normal identity for floor function and fractional part function  i.e $x = [x]  +   {x}$  using this the 
expression inside the limit can be changed to   $1- (\frac{\{x\}}{x} )$ we know that  $\{x\}$ is lying between $[0,1)$ and $x\to\infty$  hence $\frac{\{x\}}{x} \to 0$ hence the limit is $1$ .  But I recommend that you stick to the sandwich theorem solution as it seems more cool to me.
For the last part  you could start by saying as  x tends to zero we calculate LHL and RHL separately . Now in for LHL the limit is same as the limit of $\frac{1}{x}$ as $x$ tends to zero which  tends to infinity  but the RHL is similar to limit of $\frac{0}{x} =0$ as RHL is not same as LHL hence limit doesn't exist.
A: Your idea of using $x-1\le\lfloor x\rfloor<x$ (the notation $\lfloor x\rfloor$ instead of $[x]$ is more common nowadays) is good.
For $x>0$ this entails
$$
\frac{x-1}{x}\le\frac{\lfloor x\rfloor}{x}<1
$$
and for $x<0$ we get instead
$$
\frac{x-1}{x}\ge\frac{\lfloor x\rfloor}{x}>1
$$
Now squeeze both for $x\to\infty$ (first one) and for $x\to-\infty$ (second one).
Hint for the third limit: if $-1<x<0$, $\lfloor x\rfloor=-1$.
A: Rewrite
$$\frac{\lfloor x\rfloor}{x}=1-\frac{\{x\}}x.$$
As the numerator of the fraction is bounded and the denominator unbounded, the expression tends to $1$ for both $-\infty$ and $\infty$.
For the limit to $0$, notice that $$x<0\implies\lfloor x\rfloor\le-1$$ and the limit on the left cannot exist.
