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

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?

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    $\begingroup$ The more common name for the $\lfloor x \rfloor$ function is the "floor" function. $\endgroup$ – Axoren Nov 27 '14 at 17:46
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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.

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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 $x \to 0^{-}$.

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