The $\frac{\sin x}x$ limit and the floor function

Question

$$\lim_{x\to0}\left\lfloor\frac{\sin x}x\right\rfloor=\,?$$ I think it should be 1, but in my class notes it's given as 0. I don't understand, because the limit without the floor brackets is 1, yet $\lfloor1\rfloor=1$.

Answer 1

For $x\not=0$ it is always the case that $|x|>|\sin x|$, so we know that \begin{equation*} \left|\frac{\sin x}{x}\right|<1 \text{ for }x\not=0 \end{equation*} Applying the floor function will give you either $0$ or $-1$. For values of $x$ near $0$, the floor is $0$.

Answer 2

It is known that $|\sin x|\leq |x|,$ for $x \in \mathbb R$ and the equality holds only when $x=0$. Therefore for $x$ close to $0$, we have that:

$$0<\frac{\sin x}{x}<1\Rightarrow \left[\frac{\sin x}{x}\right]=0$$

and so the limit follows.

Note: Posted just a second after gobucksmath.

Answer 3

In a small neighbourhood around the origin, the function $$ \frac{\sin x}{x} $$ is less than $1$ and non-negative for all $x\ne 0$. This means that for all $x\ne 0$ in that neighbourhood, we have $$ \left\lfloor\frac{\sin x}{x}\right\rfloor=0 $$ and therefore the limit is $0$.

As pointed out by Akiva Weinberger in the comments, the reason that your argument does not work is that limits are only necessarily preserved by continuous functions, and $\lfloor \cdot \rfloor$ is not continuous at $1$.

Well done for being curious about this rather than just accepting what's in your class notes!

Answer 4

We have :

$$\forall x\in[-1,0[\,\cup\,]0,1],\quad 0\le\frac{\sin(x)}{x}<1$$

and therefore :

$$\forall x\in[-1,0[\,\cup\,]0,1],\quad \lfloor\frac{\sin(x)}{x}\rfloor=0$$

This implies that :

$$\lim_{x\to0}\lfloor\frac{\sin(x)}{x}\rfloor=0$$

Answer 5

Using the Maclaurin series of $\sin x$ we get -

$$\frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac {x^4}{5!}-\frac{x^6}{7!} ....$$

Clearly $$\frac{x^2}{3!}-\frac {x^4}{5!}+\frac{x^6}{7!} ....>0$$ since $|x^2|<1$ the numerators get smaller and smaller while the denominators keep on increasing(can be seen by taking terms in pairs). From this, we get

$$\frac{\sin x}{x} \rightarrow 1^{-} $$ as $x\rightarrow 0$ from either side. Therefore $$\lim_{x \rightarrow0}\bigg\lfloor\frac{\sin x}{x}\bigg\rfloor = 0$$