Existence of the limit of level sets (or zero set) of a continuous function Suppose I have 2 functions, $f,g: R^n\rightarrow R^n$ (happy to fill in further details as needed), and a parameter $\delta \geq 0$.
I am interested in the the sets
$$Z_\delta=\{x\, |\, f(x)-\delta g(x)=0\}$$
Specifically, I would like to show that
$ \lim_{\delta \rightarrow 0} Z_\delta$ exists. (I do realize this limit is not necessarily identical to $Z_0$. That's fine.)
How could I do this? (Or, get started doing this?)
Some potentially helpful facts:
$f,g$ are continuous/real analytic.
$Z_\delta$ are non-empty for all $\delta>0$.
$Z_\delta\subset Y \subset R^n$ for all $\delta>0$, where  $Y$ is some bounded set (independent of $\delta$).
I can provide/check further properties if it helps.
 A: This is almost certainly less useful than you are hoping. The result will depend on exactly how you define a limit of sets, but since both $f$ and $g$ are continuous, any practical definition will end up with the limit being between $Z_0 = f^{-1}(0)$ and $f^{-1}(0) \cap g^{-1}(0)$.
Obviously if both $f(x) = 0$ and $g(x) = 0$, then $x\in Z_\delta$ for all $\delta$. So those points would be in any reasonable limit of sets. However,

*

*if $f(x_0) \ne 0$, then $x_0$ has some neighborhood where $|f(x)| > \frac{|f(x_0)|}2$. Further, this neigborhood can be chosen to be compact. Since $g$ is continuous, it is bounded on the neighborhood. For small enough $\delta, |f(x) - \delta g(x)|$ is bounded away from $0$. So any reasonably limit set would not include $z_0$. as it has a neighborhood that does not intersect $Z_\delta$ for any sufficiently small $\delta$.  Hence your limit set must be a subset of $Z_0$.

*If $g(x_0) \ne 0$, there is exactly one $\delta = \frac{f(x_0)}{g(x_0)}$ for which $x_0 \in Z_\delta$. So any concept of set limit that requires $x$ actually appear in $Z_\delta$ sets for small $\delta$ to be in the limit will not include such $x_0$. For any such concept, $$\lim_{\delta \to 0} Z_\delta = f^{-1}(0)\cap g^{-1}(0)$$

*If $f(x_0) = 0$ and $g(x_0) \ne 0$, then because both are continuous, there is a neighborhood of $x_0$ where $|f(x)| < \epsilon$ and $|g(x)| > m$ for any $\epsilon, m > 0$. That is, for any neighborhood of $x_0$ and any $\delta_0 > 0$, there exists a $\delta < \delta_0$ such that $Z_\delta$ intersects the neighborhood. So there are sequences $\delta_i, a_i \in Z_{\delta_i}$ such that $\delta_i \to 0$ and $a_i \to x_0$. And therefore there is a at least one sense of "limit of sets" for which $$\lim_{\delta \to 0} Z_\delta = Z_0$$
Some particular set limits are

*

*$$\liminf_{\delta \to 0} Z_\delta = \bigcup_{\epsilon > 0}\bigcap_{0 < |\delta| < \epsilon} Z_\delta = f^{-1}(0) \cap g^{-1}(0)$$

*$$\limsup_{\delta \to 0} Z_\delta = \bigcap_{\epsilon > 0}\bigcup_{0 < |\delta| < \epsilon} Z_\delta = f^{-1}(0) \cap g^{-1}(0)$$

*$$\overline{\liminf_{\delta \to 0}}\ Z_\delta = \bigcup_{\epsilon > 0}\overline{\bigcap_{0 < |\delta| < \epsilon} Z_\delta} = f^{-1}(0) \cap g^{-1}(0)$$

*$$\overline{\limsup_{\delta \to 0}}\ Z_\delta = \bigcap_{\epsilon > 0}\overline{\bigcup_{0 < |\delta| < \epsilon} Z_\delta} = Z_0$$
$\overline\liminf_{\delta \to 0} Z_\delta = \liminf_{\delta \to 0} Z_\delta$ because all of the $Z_\delta$ are closed.
