# Eliminate removable discontinuities of non-rational function

I have a function that has a removable discontinuity at $x=0$:

$$f(x)=\begin{cases} \frac{18x^3}{(9x^2-1)+(3x^2+1)^{3/2}}, & x \ne 0 \\ 0, & x=0 \end{cases}$$

By playing around (expanding by $(9x^2-1)-(3u^2+1)^{3/2}$), I could reformulate this in a form that has two discontinuities at other places:

$$f(x)=\begin{cases} \frac{2x\left(1-9x^2+(3x^2+1)^{3/2}\right)}{3(x^2-1)^2}, & x\not\in\{-1,1\} \\ \frac{9}{8}x,& x\in\{-1,1\} \end{cases}$$

I'm not familiar with methods for "planfully" eliminating discontinuities in a term of a non-rational functions comparable to rational functions where one can simply cancel the discontinuity. Are there actually such methods, and if so, how could a single term describing this specific function be found?

• is that $u^2$ or $x^2$ in the denominator Dec 16, 2013 at 3:48
• @TrafalgarLaw: Oh, right, should be $x$! Thanks. Dec 16, 2013 at 4:03
• this function will be continuous everywhere , the only point you need to check is at $x=0$ Dec 16, 2013 at 4:08
• @TrafalgarLaw: Yes, that's clear. My motivation for this question was whether one there were methods for finding terms that do not have discontinuities if discontinuities in the original term are in fact removable. Dec 16, 2013 at 4:12

For the sake of completeness , I am showing you an example of removable discontinuity :

$$f(x)=\begin{cases} x^2, & x < 1 \\ 0, & x=1\\ 2-x,&x>1 \end{cases}$$

This shows that the LHL and RHL of the $f(x)$ are same at $x=1$ but the value of $f(x)$ at $x=1$ is $0$,if I could change the value of $f(x)$ at $1$ to $1$ then it would be continuous everywhere in its domain.

Now coming back to the question . The limit of $f(x)$ int this case exists at $0$ . Find it by applying L'Hospitals Rule you will get the value to be $1.33333$ but the value $f(0)=0$. So it is a removable discontinuity at $0$.