Find $\lim_{x \to -1} 1/(\sqrt{|x|-\{-x\}})$ where $\{\}$ denotes the fractional part.

Question

Find $\lim_{x \to -1} 1/(\sqrt{|x|-\{-x\}})$ where $\{\}$ denotes the fractional part.

My attempt - $$\lim_{x \to -1} \frac{1}{\sqrt{|x| -\{x\}+1}}\\ \lim_{x \to -1} \frac{1}{\sqrt{|x|+1 -x +[x]}}$$

Now, $$\lim_{x \to -1+} \frac{1}{\sqrt{1+1+1 -1}} = \frac{1}{\sqrt{2}}\\ \lim_{x \to -1-} \frac{1}{\sqrt{1+1+1-2}} = 1$$

So, limit doesn't exist. But answer is given, that limit exists and is equal to $1$. Where did I go wrong$?$

Answer 1

If $x\in(-2,-1]$, then $|x|-\{-x\}=-x-(-1-x)=1$.

If $x\in(-1,0]$, then $|x|-\{-x\}=-x-(-x)=0$.

So the left-sided limit exists and equals $1/\sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.

Answer 2

In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$\lim_{m\to1}\frac 1{\sqrt{m-\{m\}}}=\lim_{m\to1}\frac 1{\sqrt{\lfloor m\rfloor}}$$Now note that $[0,1)$ does not belong to the domain of $\dfrac1{\sqrt{\lfloor m\rfloor}}$, so only one sided-approach, $m\to1^+$, is possible. We don't require $m\mapsto\dfrac1{\sqrt{\lfloor m\rfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $\lim_{x\to0}\sqrt x=0$, even though $x\mapsto\sqrt x$ is not defined for $x<0$. Thus, the answer is$$\lim_{m\to1}\frac 1{\sqrt{\lfloor m\rfloor}}=\lim_{h\to0^+}\frac 1{\sqrt{\lfloor1+h\rfloor}}=1$$.

Answer 3

In the negatives, $|x|=-x$ so that $-x-\{-x\}=\lfloor-x\rfloor$. So the function values on the left and on the right are

$$\frac1{\sqrt1},\color{red}{\frac1{\sqrt0}}$$

and the limit is $1$ (as a limit is computed inside the domain).