Given a function $f$ defined by $$f(x) := \left( x^\sigma + b \right)^{1/\sigma}, \qquad (\sigma <0, \quad b \in \mathbb{R}_+).$$

Since $\sigma <0$ and thus the negative exponent (power) functions $x^\sigma$ (or $x^{1/\sigma}$) are not well defined for $x=0$, it seems like this function $f$ is also not well defined for $x=0$.

However, when I draw the graph of this function $f$, I found the graph of $f$ could reach $(0,0)$ (i.e., $f(0)=0$), so that I am confused whether such a function $f$ is well defined on zero point.

Question 1: It is clear that the above function $f$ is well defined on $(0, \infty)$, but I am wondering that is $f$ well defined on $x =0$ point?

Question 2: If $f$ is well defined at $x=0$ (and hence well defined on $\mathbb{R}_+$), then is this function $f$ continuous at $x=0$? If so, how to prove it?

Any idea or suggestions are most welcome and much appreciated!

Thank you in advance!


$f(0)$ is not defined because you cannot calculate $0^\sigma$ as you say. However you can calculate $\lim_{x\to 0^+}f(x)$. Intuitively, as $x$ gets very small $x^\sigma$ will get very large so $b$ will not matter. You will then have $f(x)$ very close to $x$ for small $x$ and the limit will be $0$. If you want, you could define a new function $g(x)$ by $$g(x)=\begin {cases} 0&x=0\\f(x)&x \gt 0 \end {cases}$$ Now $g(x)$ is well defined and continuous from above at $0$. It is the same idea as a removable singularity in a function, where you "fill in a hole" of the definition in such a way that you make the function continuous.

  • $\begingroup$ Thank you very much @Ross Millikan :-) It's a huge help to me. I learned it. $\endgroup$ – Paradiesvogel Aug 29 '18 at 23:25
  • $\begingroup$ Dear @Ross Millikan, I have a question that for $\sigma <0$, what if we define $0^\sigma = +\infty$ and $\infty^{\sigma} =0$ in advance? Then under this setting, is the function $f = (x^\sigma + b)^{1/\sigma}$ well defined on $\mathbb{R}_+$ and continuous on $\mathbb{R}_+$ everywhere? Thanks so much again :-) $\endgroup$ – Paradiesvogel Aug 31 '18 at 3:33
  • $\begingroup$ In the reals raising negative numbers to non-integral powers is not defined. What would you think $(-2)^{1/\pi}$ is? In the complex plane it is defined but multivalued so $f(x)$ is not defined for negative $x$ unless $\sigma$ is a negative integer. $\endgroup$ – Ross Millikan Aug 31 '18 at 3:47
  • $\begingroup$ Thanks @Ross Millikan . I just want to focus on the domain of $f$ to be non-negative rather than considering negative $x$, and I also set the constant $b$ to be non-negative. In fact, I am curious and wondering that under a convention that $0^\sigma := +\infty$ and $(+ \infty)^\sigma := 0$ for $\sigma <0$, is the function $f$ well defined on $[0, \infty)$ and also continuous at $x = 0$? Your kind help is much appreciated! :-) Thank you $\endgroup$ – Paradiesvogel Aug 31 '18 at 3:53
  • $\begingroup$ You need $x^\sigma+b \gt 0$ for the $1/\sigma$ power to make sense. You need $x \gt 0$ for the $x^\sigma$ to make sense, so as long as $b \ge 0$ your domain is $x \gt 0$ and $f$ will be continuous there. With the addition of $f(0)=0$ you get the domain to be $x \ge 0$. You don't need what you suggest in the last comment for that. $\endgroup$ – Ross Millikan Aug 31 '18 at 4:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.