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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 $x\to 0$.

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@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
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6 Answers

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.

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[x] is the ceiling of x –  eva xxx Dec 26 '10 at 20:40
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What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0<x<1$.

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

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:

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

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

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⌈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
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@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
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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.

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

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

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

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

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