What to do when we have a fractional part of $x$?

Question

How to find

$$\lim_{x\rightarrow 0} \frac{\{x\}}{\tan\{x\}}$$

Where $\{x\}$ is the fractional part of $x$.

Answer 1

Note that $$\lim_{x\to 0^+}\{x\}=0$$ and $$\lim_{x\to 0^-}\{x\}=1$$ since $-1<x<0$ implies $\{x\}=x-\lfloor x \rfloor=x-(-1)=x+1$. Then $$\lim_{x\to0^+}\frac{\{x\}}{\tan\{x\}}=\lim_{t\to 0}\frac t{\tan t}$$ and $$\lim_{x\to0^-}\frac{\{x\}}{\tan\{x\}}=\lim_{t\to 1}\frac t{\tan t}$$

Answer 2

The limit from the right side is easy. Notice that for $x\in(0,1)$, $\{x\}=x$, thus

$$\lim_{x\to0^+}\frac{\{x\}}{\tan\{x\}}=\lim_{x\to0^+}\frac x{\tan(x)}=\lim_{x\to0^+}\cos(x)\frac x{\sin(x)}=\cos(0)\times1=1$$

Which follows nicely from the famous $\lim_{x\to0}\frac{\sin(x)}x=1$ limit.

Now notice that for $x\in(-1,0)$, $\{x\}=x+1$, thus

$$\lim_{x\to0^-}\frac{\{x\}}{\tan\{x\}}=\lim_{x\to0^-}\frac{x+1}{\tan(x+1)}=\frac1{\tan(1)}$$

Since the left and right limits don't agree... the limit doesn't exist.