# Eliminate removable discontinuity

I have two rational functions which I have to examine for discontinuity and try to remove their domain gaps if possible.

$$f(x)=\frac{|x|-1}{x^3-x}$$ and $$g(x)=\frac{sin (x)}{sin (2x)}$$

I determined the domain of the first function: $\mathbb{D}=\mathbb{R}\backslash \{-1,0,1\}$, and I also simplified it to a form: $f(x)=\frac{1}{x(x+1)}$ I tried to determine the limits for the undefined values next. $$\lim\limits_{x \rightarrow 0}= \frac{1}{0(0+1)}=undefined$$ $$\lim\limits_{x \rightarrow 1}= \frac{1}{1(1+1)}=\frac{1}{2}$$ $$\lim\limits_{x \rightarrow -1}= \frac{1}{-1(-1+1)}=undefined$$

After removing the possible gaps my function looks like: $$f(x)=\begin{cases}\frac{|x|-1}{x^3-x}&if x\neq0,x\neq-1\\ \frac{1}{2}&if x=1\\\end{cases}$$

Now, for the second function $g(x)$ I determined the domain which is: $\mathbb{D}=\mathbb{R}\backslash\{\frac{\pi n}{2}, n \in \mathbb{Z}\}$ But I can't really find a way to simplify this function, and also all of the limits seem to be undefined. Does it mean I can't remove the gaps here?

I just need to know if my approach to solve this kind of a problem is correct, or did I miss something?

• $\sin(2x)$ or $\sin^2(x)$ ?
– user65203
Nov 30, 2015 at 16:52

$$x^2-1=(|x|+1)(|x|-1),$$ hence you simplify as $$\hat f(x)=\frac1{x(|x|+1)},$$ which is only undefined at $x=0$.

Also, $$\sin(2x)=2\sin(x)\cos(x)$$ and $$\hat g(x)=\frac1{2\cos(x)}.$$

$f(x)$ does not simplify to a form $\frac{1}{x(x+1)}$. It simplifies into the form $$\frac{1}{x(x+1)} \cdot \frac{|x|-1}{x-1}$$

which is certainly not the same!

For the second, try to use the fact that $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$

and multiply the whole function by $\frac{x}{x}$.

• Can the downvoter explain why this answer is bad and deserving of a downvote?
– 5xum
Nov 30, 2015 at 17:03
• But how can I remove the discontinuity if the $$f(x)=\frac{1}{x(x+1)}*\frac{|x|-1}{x-1}$$ I should be able to remove the discontinuity if the numerator and the denominator equals 0. Which in this case should be for 1 or -1, but if I try -1 or 1 as a value of x the equation becomes just undefined. Nov 30, 2015 at 17:14
• @MonsieurMolly You remove the discontinuity by calculating the limit. When calculating the limit of $x\to -1$, you can for example know that $|x|=-x$, which will greatly simplify the calculation. Similarly for $x\to 1$, when $|x|=x$. For $x\to 0$, look at the left and right limit separately.
– 5xum
Nov 30, 2015 at 17:21
• I guess i don't get it like totally right. I just tried to calculate the limit for $x \rightarrow 1$ Which ist $\frac{1}{2}$ and also calculated the limit for $x\rightarrow -1$ which is undefined, but why? I get something like this when I substitute $\lim\limits_{x \to -1}{\frac{1}{x(x+1)}*\frac{-x-1}{x-1}}=\frac{1}{-1(-1+1)}*-1$ which is again undefined. On the other hand wolfram alpha says $\lim\limits_{x\to -1}=-\frac{1}{2}$ Tell me... should I calculate the limits for 1 and -1 separately on the both sides for each one i.e $\lim\limits_{x\to -1^+}$ and $\lim\limits_{x\to -1^-}$? Nov 30, 2015 at 18:42
• What's the relevance of the second hint ?
– user65203
Nov 30, 2015 at 19:21