1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ is that $u^2$ or $x^2$ in the denominator $\endgroup$
    – abkds
    Dec 16, 2013 at 3:48
  • $\begingroup$ @TrafalgarLaw: Oh, right, should be $x$! Thanks. $\endgroup$
    – Thomas W.
    Dec 16, 2013 at 4:03
  • $\begingroup$ this function will be continuous everywhere , the only point you need to check is at $x=0$ $\endgroup$
    – abkds
    Dec 16, 2013 at 4:08
  • $\begingroup$ @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. $\endgroup$
    – Thomas W.
    Dec 16, 2013 at 4:12

1 Answer 1

1
$\begingroup$

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

enter link description here

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

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 $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .