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In a formal mathematics context does the statement $$\lim_{f(x)\to a}k(x)$$ where $f(x)\neq c$, where $c$ is a constant, make sense? For example does $$\lim_{x^2\to 0}x$$ make any sense in a formal and rigorous setting.

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  • $\begingroup$ Edited, I excluded the possibility of constant functions. $\endgroup$ – Will Fisher Jul 26 '16 at 21:54
  • $\begingroup$ I suppose if $f$ is invertible, it wouldn't be unreasonable to define $$\lim_{f(x) \to a} k(x) = \lim_{y\to a} k(f^{-1}(y))$$ but what is the utility? If $f$ isn't invertible, it makes no sense: e.g. what is $$\lim_{(x^2-1) \to 0} x?$$ $\endgroup$ – User8128 Jul 26 '16 at 21:54
  • $\begingroup$ With some tweaking, this is a good, natural question. As @User8128 notes, a necessary condition for this to make good sense would be that $f$ is locally invertible everywhere. Even then, in general topological spaces, it is not true that continuous + invertible implies that a function is open. In situations where that does hold, then, yes, your $f$ has at least a local inverse, etc. $\endgroup$ – paul garrett Jul 26 '16 at 22:04
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It's not quite standard notation, but the meaning is pretty clear: $$ \lim_{f(x) \to a} k(x) = B$$ would mean:

For every $\epsilon > 0$ there exists $\delta > 0$ such that whenever $0 < |f(x) - a| < \delta$, $|k(x) - B| < \epsilon$.

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For every $c>0$ there exists $d>0$ such that $|f(x)-a|<d$ implies that $k(x)-l|<c$

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