8
$\begingroup$

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 $x\to 0$.

  • 0
    You know, this question has made me acquire a deep irrational, $\pi$-like hatred for limits.2011-06-06

6 Answers 6

9

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.

  • 2
    [x] is the ceiling of x2010-12-26
8

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

4

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

-1

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

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:

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

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

  • 2
    @sigma - When you get 0/0 from direct substitution when doing limits, it not undefined, it's indeterminate. The limit may still exist.2011-06-06
-1

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.

-1

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

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

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