How to calculate this limit $?$ :
$$ \lim_{x\ \to\ 0^{+}}\,\,\,\frac{x}{a} \left\lfloor\,{\frac{b}{x}}\,\right\rfloor\,,\qquad a, b > 0 $$ where $\left\lfloor\,{x}\,\right\rfloor$ represents $greatest\ integer\ function$ or floor function, i.e greatest integer less than or equal to $x$.
We can start off by substituting $\frac{b}{x}$ $$L=\lim_{x \rightarrow 0^+}\frac{x}{a}\left\lfloor \frac{b}{x} \right\rfloor=\lim_{x \rightarrow 0^+}\frac{bx}{a}\left\lfloor \frac{1}{x} \right\rfloor=\frac{b}{a}\lim_{x \rightarrow \infty}\frac{\lfloor x\rfloor}{x}$$ Intuitively, the limit on the right-hand side approaches 1 because the difference between $x$ and $\lfloor x\rfloor$ becomes negligible as $x$ approaches infinity, but we can prove it using algebra $$L=\frac{b}{a}\lim_{x \rightarrow\infty}\frac{x-[x]}{x}=\frac{b}{a}\left(1-\lim_{x\rightarrow\infty}\frac{[x]}{x}\right)$$ where [x] is the fractional part of x
The fractional part of x, [x], has a range $0≤y≤1$, so we can use the squeeze theorem. Keep in mind we can prove this using the exact same methods for $x\rightarrow -\infty$ $$\lim_{x\rightarrow\infty}\frac{0}{x}≤\lim_{x\rightarrow\infty}\frac{[x]}{x}≤\lim_{x\rightarrow\infty}\frac{1}{x}$$ $$0≤\lim_{x\rightarrow\infty}\frac{[x]}{x}≤0$$ $$\lim_{x\rightarrow\infty}\frac{[x]}{x}=0$$ Plugging it back in we get $$L=\frac{b}{a}\left(1-\lim_{x\rightarrow\infty}\frac{[x]}{x}\right)=\frac{b}{a}(1-0)=\frac{b}{a}$$ $$L=\lim_{x \rightarrow 0^+}\frac{x}{a}\left\lfloor \frac{b}{x} \right\rfloor=\frac{b}{a}$$