Value of $\Bigl\lfloor\,\lim_{x\to0}\frac{\sin x}{x}\Bigr\rfloor$

Question

What is the value of $\Bigl\lfloor\,\lim_{x\to 0}\frac{\sin x}{x}\Bigr\rfloor$? Is it $1$ or $0$?

I was told that the answer is $0$ as $\sin{x}$ is less than $x$ as $x\rightarrow0$. Is it correct or are limits exact values?

I know that $\lim_{x\to0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor$ will be $0$ due to the above mentioned fact.

Answer 1

We have $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{\cos x}{1}=\cos (0) = 1.$$ Since $\lfloor 1\rfloor=1$ it follows that $\left\lfloor\left(\lim_{x\to 0}\left(\frac{\sin x}{x}\right)\right)\right\rfloor=1$.

Answer 2

By the Maclaurin's expansion of $sin(x)$, we have, $$\sin(x) =\sum^{\infty}_{k=0}\frac{(-1)^k}{(2k+1)!}x^{2k+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

As $x\to0$ we compute left hand limit, $$\lim_{x\to0^{-}}\frac{\sin(x)}{x} =\lim_{x\to0^{-}}\left( 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right) $$ $$\lim_{x\to0^{-}}\frac{\sin(x)}{x} = 1$$ Further for right hand limit, $$\lim_{x\to0^{+}}\frac{\sin(x)}{x} =\lim_{x\to0^{+}}\left( 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right) $$ $$\lim_{x\to0^{+}}\frac{\sin(x)}{x} = 1$$ Now, $$\Bigl\lfloor\lim_{x\to0^{+}}\frac{\sin(x)}{x}\Bigr\rfloor = \Bigl\lfloor\lim_{x\to0^{-}}\frac{\sin(x)}{x}\Bigr\rfloor =1 $$

Answer 3

Since

• $\lim_{x\to0}(\frac{\sin x}{x})=1 $
• and for $x\ne 0$ sufficiently small: $0<\frac{\sin x}{x}<1$

we have that

$$\Bigl\lfloor\,\lim_{x\to0}\frac{\sin x}{x}\Bigr\rfloor=1$$

and

$$\lim_{x\to0}\Bigl\lfloor\frac{\sin x}{x}\Bigr\rfloor=0$$