What is $\displaystyle \lim_{x \to 0} \frac{\lceil x \rceil}{x}$ ? Here, $\lceil x \rceil$ is the ceiling function at $x$.
For left limit and right limit as $x\to 0$.
What is $\displaystyle \lim_{x \to 0} \frac{\lceil x \rceil}{x}$ ? Here, $\lceil x \rceil$ is the ceiling function at $x$.
For left limit and right limit as $x\to 0$.
Assuming that [x] is the floor of x, then look at this graph.
Assuming that [x] is the ceiling of x, then look at this similar looking graph.
Assuming that [x] is the "nearest integer" function, then consider what the nearest integer is on the interval [-0.49, 0.49].
If this is something else, please specify.
No need for graphs or WA. Just note that
$$ \lceil x \rceil = \left\{ \begin{array}{cl} 1 & x \in (0,1) \\ 0 & x \in (-1,0) \end{array} \right.$$
Hence, $\displaystyle\lim_{x \to 0^+} \frac{\lceil x \rceil}{x} = \lim_{x \to 0^+} \frac{1}{x} = + \infty$. On the other hand, $\displaystyle\lim_{x \to 0^-} \frac{\lceil x \rceil}{x} = \lim_{x \to 0^-} \frac{0}{x} = 0$.
What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0
$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$
This is what the Alpha Wolf says:
$\lim \limits_{x \to 0^-} \frac{\lceil x \rceil}{x} = 0$
$\lim \limits_{x \to 0^+} \frac{\lceil x \rceil}{x} = \infty$
The two limits aren't equal. And what could this possible mean?
How did we get this? Look here:
$\lim \limits_{x \to 0^-} \lceil x \rceil = 0$
$\lim \limits_{x \to 0^+} \lceil x \rceil = 1$
And:
$\lim \limits_{x \to 0^-} x = 0$
$\lim \limits_{x \to 0^+} x = 0$
The limit is equivalent to the derivative of the ceiling function at 0. The derivative of the ceiling function, by observing the graph, is constant when x is not an integer, and undefined when x is an integer. The above limit is therefore undefined.
$$\lim_{x \to 0^+} \frac{\lceil x \rceil}{x} = \frac{\lim_{x \to 0^+}\;\lceil x \rceil}{\lim_{x \to 0^+} \;x} = \frac{1}{\lim_{x \to 0^+} \;x} = \lim_{x \to 0^+}\frac{1}x = \infty$$
$$\lim_{x \to 0^-} \frac{\lceil x \rceil}{x} = \frac{\lim_{x \to 0^-}\;\lceil x \rceil}{\lim_{x \to 0^-} \;x} = \frac{0}{\lim_{x \to 0^-} \;x} = \lim_{x \to 0^-}\frac{0}x = 0$$
$$\lim_{x \to 0^+} \frac{\lceil x \rceil}{x} \not = \lim_{x \to 0^-} \frac{\lceil x \rceil}{x} \therefore \;\, \not \exists \;\;\lim_{x \to 0} \frac{\lceil x \rceil}{x}$$