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For any given limit value $L$, there are uncountably many functions such that $\displaystyle\lim_{x \to c}f(x) = L$.
To see this, pick any one function such that $\displaystyle\lim_{x \to c}f(x) = L$, and define a new function $\bar{f}$ such that $\bar{f}(x) = f(x)$ for all $x$ except $\{x_1,\ldots,x_n\}$ (a finite set of points). You can define $\bar{f}(x_1), \ldots, \bar{f}(x_n)$ arbitrarily. Then, $\displaystyle\lim_{x \to c}\bar{f}(x) = L$. Of course, there are more functions whose limit as $x \to c$ is $L$.
Sticking to "nice" functions, you can still come up with examples of functions with the same limit. For example, $\displaystyle\lim_{x \to 0}e^x = \lim_{x \to 0}\dfrac{\sin x}{x} = \lim_{x \to 0}(x^2+1) = \lim_{x \to 0}1 = 1$.
Thus, if you are given the value of a limit, you cannot uniquely determine what the function was. Hence, there is no "inverse limit" operation.
Such an operation may not exist, as a limit is not a bijective "mapping"
Just as binary multiplication takes $$ \cdot \,\, : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$, as in $2 \cdot 3 = 6$, we may not recover this information through any sort of inverse operation, bijectively. That is, given the number $6$, I cannot construct
$$A: \mathbb{R} \to \mathbb{R} \times \mathbb{R}$$ for this is clearly not bijective. $A(6) = 2 \cdot 3, 6 \cdot 1, 5 \cdot \frac{6}{5}$, etc. The same sort of issue exists when trying to take an inverse limit operation.
$$\lim_{x \to \infty} x^n = \lim_{x \to \infty} \sum_{n \in N} x^n = \, \, \text{etc} \,\, = \infty$$