# Why is $\lim_{x\to 0^+} \left\lfloor\frac{\sin|x|}{x}\right\rfloor= 0\,?$

Problem: Check $$\lim_{x\to 0} \left\lfloor\frac{\sin|x|}{x}\right\rfloor$$

Now, what the book does is this:

$$\textrm{RHL} \implies\lim_{x\to 0^+} \left\lfloor\frac{\sin|x|}{x}\right\rfloor = \lim_{h\to 0} \left\lfloor\frac{\sin|0 +h|}{0+h}\right\rfloor$$

We know, $\dfrac{\sin h}{h}\to 1$ as $h\to 0$ but less than 1

$$\therefore \textrm{RHL}= 0$$

Again, $$\textrm{LHL}\implies \lim_{h\to 0} \left\lfloor\frac{\sin|0 -h|}{0-h}\right\rfloor$$

We know $\dfrac{\sin h}{-h}\to -1$ as $h\to 0$ but greater than -1

$$\therefore \textrm{LHL}= -1$$

I really couldn't conceive what the author meant by those bold phrases.

• Could the second bold phrase be "but greater than -1" Notice the minus... ? – GambitSquared Feb 11 '16 at 15:44

What the author is saying is that, when $h>0$, $$\frac{\sin h}h<1,\ \ \ \text{ so } \left\lfloor\frac{\sin h}h\right\rfloor=0.$$
In the second part he says that, if you approach from the left, $$\frac{\sin h}{-h}>-1,\ \ \ \text{ so } \left\lfloor\frac{\sin h}h\right\rfloor=-1.$$
• And in other words.. it means that $\frac{\sin h}{-h}$ approaches $-1$ when $h-->0$ from the right side, and $\frac{\sin h}{h}$ approaches $1$ when $h-->0$ from the left side. – Sijaan Hallak Feb 11 '16 at 15:50