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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!

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$f(0)$ 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.

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  • $\begingroup$ 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

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