$$ \lim_{x \to 0} \frac{a}{x} \left\lfloor\frac{x}{b} \right\rfloor $$
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?
On
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 $x \to 0^{-}$.
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.