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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \lim_{x \to 0} \frac{a}{x} \left\lfloor\frac{x}{b} \right\rfloor $$

The $\lfloor \rfloor$ stands for the greatest integer function.

I have calculated and the left-hand limit is coming as (ab). But, I have doubt in the right-hand limit. I did this problem by sandwich-theorem. Can, anyone help me to find the right-hand limit correctly?

share|cite|improve this question
The more common name for the $\lfloor x \rfloor$ function is the "floor" function. – Axoren Nov 27 '14 at 17:46

Hint: Assuming $b \ne 0$,

$$ \lim_{x \to 0} \frac{a}{x} \left\lfloor\frac{x}{b} \right\rfloor = \lim_{x \to 0} \frac{a}{bx} \left\lfloor \frac{bx}{b}\right\rfloor = \lim_{x \to 0} \frac{a}{b} \left( \frac{\lfloor x\rfloor}{x}\right) = \frac{a}{b} \lim_{x \to 0} \frac{\lfloor x\rfloor}{x} $$

For $x$ near $0$, $\lfloor x \rfloor$ is just $0$ on the right and $-1$ on the left.

share|cite|improve this answer

I assume that $a > 0$ and $b > 0$.

If $0 < x < b$, $\lfloor \frac{x}{b} \rfloor = 0$.

Therefore, for $0 < x < b$, $\frac{a}{x}\lfloor \frac{x}{b} \rfloor = 0$.

Therefore, $\lim_{x \to 0^{+}}\frac{a}{x}\lfloor \frac{x}{b} \rfloor = 0$.

The expression is not defined at $x = 0$.

If $0 > x > -b$, $\lfloor \frac{x}{b} \rfloor = -1$.

Therefore, for $0 > x > -b$, $\frac{a}{x}\lfloor \frac{x}{b} \rfloor = -\frac{a}{x}$.

This is not defined as $x \to 0^{-}$.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.