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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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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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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What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0<x<1$. |
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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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$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$ This is what the Alpha Wolf says:
The two limits aren't equal. And what could this possible mean? How did we get this? Look here:
And:
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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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