Evaluate $\lim_{x \to 0} \left\lfloor(1-e^x)\frac {\sin x}{|x|}\right\rfloor$

Question

How to simplify the following limit:$$ \lim_{x \to 0} \left\lfloor(1-e^x)\frac {\sin x}{|x|}\right\rfloor,$$ where $\lfloor\cdot\rfloor$ represents the greatest integer function.

Given limit is in 0/0 form so can we apply
LH rule to solve this problem.. or there is special approach for the same.

Answer 1

Since, when $x$ is close enough to $0$, $$\left\lfloor(1-e^x)\frac{\sin x}{\lvert x\rvert}\right\rfloor=-1,\tag1$$your limit is equal to $-1$.

The fact that $(1)$ holds when $x$ is close enough to $0$ can be deduced from these facts:

• if $x$ is close to $0$ and $x>0$, then $0<\frac{\sin x}x=\frac{\sin x}{\lvert x\rvert}<1$. But $-1<1-e^x<0$, and therefore $(1-e^x)\frac{\sin x}{\lvert x\rvert}\in(-1,0)$;
• if $x$ is close to $0$ and $x<0$, then $0<\frac{\sin x}x<1$, and therefore $-1<\frac{\sin x}{\lvert x\rvert}<0$. Besides, $0<1-e^x<1$, and therefore $(1-e^x)\frac{\sin x}{\lvert x\rvert}\in(-1,0)$ too.