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Let $f(x)=x \cdot \sqrt {x-1}$

Normally, do you consider $f(0)=0$ or undefined?

Thanks in advance

Ahmed Ali
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    i would argue it is undefined because (without complex numbers) you can't take the squareroot of a negative number. You can not just argue zero times anything is zero if that "anything" breaks the laws of domains, however, if you bring the x term inside the radical, then x=0 is part of the domain – imranfat May 06 '14 at 18:09
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    Depends on the domain and codomain of $f$. – FH93 May 06 '14 at 18:09
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    Even the limit $\lim_{x\rightarrow 0}f(x)$ is not real, so if $f$ is defined as a real function then it is undefined at $0$. – Samrat Mukhopadhyay May 06 '14 at 18:10
  • @imranfat I tend to agree with you. I was shocked to find a similar function defined to equal $0$ in Leithold's textbook. – Ahmed Ali May 06 '14 at 18:20
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    Samrat: I disagree with you. I would argue that if you allow $f$ to be complex-valued then the limit equals zero, which is real. – JPi May 06 '14 at 18:25
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    @AhmedAli On a side note here, this example clearly illustrates one has to be careful with "factoring out" an x from the square root. It does change the domain of the resulting function in this case. Had the "x" been inside, a single point (0,0) would be part of the domain – imranfat May 06 '14 at 19:05

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If $f$ it's a real function than its domain is $x\ge1$, which doesn't include $0$, so $f(0)$ is not defined

Alessandro Codenotti
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