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I'm trying to find the nulls of $f(x) = \frac{x}{\sqrt{x^2+x}}$. I'd claim $f(x) \neq 0$ because it's not part of the domain. However by transforming: $f(x) = \sqrt{\frac{x}{x+1}}.$ So it can become $0$ zero somehow. However, does my claim still hold? In my view $f(x) = 0$ just if $x\to0$. I'm not right sure though. How to call such kind of point?

Leon
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  • You claim to have written $f(x)$ twice, but the two representations have different domains, so are not the same function. – Eric Towers Apr 24 '21 at 22:06
  • I've just made some transformations. Basically it's the same functions with different domains. That's weird. – Leon Apr 24 '21 at 22:12
  • There is no such that as "the same function with different domains". Specification of domain is part of a function. If you change the domain, you get a different function. – Eric Towers Apr 24 '21 at 22:14
  • The function has the same domain for x>0 – Тyma Gaidash Apr 24 '21 at 22:47
  • my essential question is: is $f(x)\neq 0$ ? Is $0$ truly a limit? – Leon Apr 24 '21 at 22:54
  • The thing is: can u claim something is not $0$ if it's not defined – Leon Apr 24 '21 at 22:59
  • @Leon this can be called “by convention” and you can prove that the lim of the first function as x approaches 0 is 0 by differentiating the top to get $\frac{0}{\frac{2x+1}{2\sqrt{x^2+1}}}$=0 – Тyma Gaidash Apr 24 '21 at 22:59

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