Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x\to 0$.

share|improve this question
@eva You may want to pick an answer as best. Just click on the check mark on the answer. –  muntoo Jan 7 '11 at 6:20
You know, this question has made me acquire a deep irrational, $\pi$-like hatred for limits. –  muntoo Jun 6 '11 at 1:07
add comment

6 Answers 6

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.

share|improve this answer
[x] is the ceiling of x –  eva xxx Dec 26 '10 at 20:40
add comment

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, $\displaystyle\lim_{x \to 0^-} \frac{\lceil x \rceil}{x} = \lim_{x \to 0^-} \frac{0}{x} = 0$.

share|improve this answer
add comment

What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0<x<1$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$

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:

$\lim \limits_{x \to 0^-} \lceil x \rceil = 0$

$\lim \limits_{x \to 0^+} \lceil x \rceil = 1$


$\lim \limits_{x \to 0^-} x = 0$

$\lim \limits_{x \to 0^+} x = 0$

share|improve this answer
⌈x⌉ is discontinuous at x = 0? –  user2468 Dec 27 '10 at 3:06
@M.S. And that means what...? –  muntoo Dec 27 '10 at 3:12
Surely you mean $\lim_{x\to 0^+} \lceil x \rceil = 1$ instead of $\infty$. –  Rahul Dec 27 '10 at 3:22
@Rahul Fixed - That's what happens when you translate from one "language" to another. :) –  muntoo Dec 27 '10 at 3:24
@sigma - When you get 0/0 from direct substitution when doing limits, it not undefined, it's indeterminate. The limit may still exist. –  jd.r Jun 6 '11 at 2:28
show 2 more comments

$$\lim_{x \to 0^+} \frac{\lceil x \rceil}{x} = \frac{\lim_{x \to 0^+}\;\lceil x \rceil}{\lim_{x \to 0^+} \;x} = \frac{1}{\lim_{x \to 0^+} \;x} = \lim_{x \to 0^+}\frac{1}x = \infty$$

$$\lim_{x \to 0^-} \frac{\lceil x \rceil}{x} = \frac{\lim_{x \to 0^-}\;\lceil x \rceil}{\lim_{x \to 0^-} \;x} = \frac{0}{\lim_{x \to 0^-} \;x} = \lim_{x \to 0^-}\frac{0}x = 0$$

$$\lim_{x \to 0^+} \frac{\lceil x \rceil}{x} \not = \lim_{x \to 0^-} \frac{\lceil x \rceil}{x} \therefore \;\, \not \exists \;\;\lim_{x \to 0} \frac{\lceil x \rceil}{x}$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.