Existence of a limit .. Does  $\lim_{x \to 0} f(x)$  exist or not?, where $f(x)= \frac{[x]}{x}$.
I know what is $[x]$. Here $[x]$ denotes the greatest integer not greater than $x$.
Please someone give some hints or solution.
 A: In order for this limit to exist, the limit as approached from the left and the limit as approached from the right must both exist and be equal to each other.
From the left, $x$ is a very small negative number going to $0$, and $\lfloor x \rfloor$ is $-1$. This means that the limit from the left is $+\infty$. A limit from the left can be written with a superscript minus sign:
$$\displaystyle\lim_{x\to0^{-}} \displaystyle\frac{\lfloor x \rfloor}{x} = \infty$$
From the right, $x$ is a very small positive number going to $0$, and $\lfloor x \rfloor$ is $0$. This means that the limit from the right is $0$. A limit from the right can be written with a superscript plus sign:
$$\displaystyle\lim_{x\to0^{+}} \displaystyle\frac{\lfloor x \rfloor}{x} = 0$$
Since these two results do not agree, the limit of $f(x)$, as $x\to0$, does not exist. 
A: To evaluate $\lim_{x \rightarrow 0}f(x)$  we have to evaluate both the left hand limit as well as the right hand limit.
We can see that  
$$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-}\frac{\lfloor x \rfloor}{x}$$
$$= \frac{-1}{-x} \rightarrow \frac{1}{0}= \infty$$
Whereas the right hand limit evaluates as $ \lim_{x \rightarrow 0^+} f(x)$ 
$$=\lim_{x \rightarrow 0^+}\frac{\lfloor x \rfloor}{x}$$
$$= lim_{x \rightarrow 0^+} \frac{0}{x}$$
$$=0$$
Since the left hand limit is not equal to the right hand limit we can conclude that the above function's limit does not exist when $x \rightarrow 0$
