$\lim_\limits{ x \to 0} (\text{fractional part of} \frac{x}{\tan(x)})$.

Question

Evaluate $\lim_\limits{ x \to 0} (\text{fractional part of} \frac{x}{\tan(x)})$. I know that as $x \to 0$, $\tan(x) \to x+$, so $\frac{x}{\tan(x)} \to 1-$, so even the fractional part $\to 1-$. But would you write the answer as $\to 1-$ or $\to 1$? I don't think you can write $1$ as the final answer because it is out of the range of the fractional part function. Thanks in advance.

Answer 1

I really do not know if this could be of any help to you.

Using Taylor series around $x=0$, we have $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ which makes $$\frac{x}{\tan(x)}=\frac x {x+\frac{x^3}{3}+O\left(x^5\right)}$$ Now, long division to get $$\frac{x}{\tan(x)}=1-\frac{x^2}{3}+O\left(x^4\right)$$

Answer 2

Even though 1 is out of range of the function "fractional part," in this case the limit would be 1, just as, strictly speaking, no value of x makes (sin x)/x equal to 1.

Answer 3

If $\lfloor z\rfloor$ denotes the largest integer less than or equal to $z$, then the fractional part of $z>0$ is $\{z\}=z-\lfloor z\rfloor$.

Let's look at the limit from the right. We know that, for $0<x<\pi/2$, we have $x<\tan x$, because the function $\tan x-x$ is strictly increasing. Therefore $0<\frac{x}{\tan x}<1$ and so $\lfloor\frac{x}{\tan x}\rfloor=0$. Thus $$ \lim_{x\to0^+}\left\{\frac{x}{\tan x}\right\}= \lim_{x\to0^+}\frac{x}{\tan x}=1 $$ You can observe that the values of $\{\frac{x}{\tan x}\}$ are (in a right neighborhood of $0$), less than $1$; somebody writes this as $1^{-}$, but I don't think it's common: the limit is $1$; the fact that the function “approaches the limit from below” can be useful in some cases.

Since the function $x/\tan x$ is even, the limit from the left is the same as the limit from the right.