Is there a way to define $\displaystyle\lim_{\varepsilon\to0}\int_{-\varepsilon}^\varepsilon\mathrm{sgn}(x)\delta(x)\,\mathrm dx$? or in general
$$\displaystyle\lim_{\varepsilon\to0}\int_{-\varepsilon}^\varepsilon f(x)\delta(x)\,\mathrm dx,\quad f = \left\{\begin{matrix}f(x) = h(x)&x<0\\f(x) = g(x)&x >0\end{matrix}\right.$$
My physics text claim it is $1$ and $g(x)$, but I'm not sure if that is math or "math", or if there are any way to get that answer?
No, such "results" should be quickly discarded by serious mathematicians. Let me begin by showing that $\lim _{\varepsilon \to 0} \int _{-\varepsilon} ^\varepsilon \mathrm{sgn}(x) \delta(x) \ \Bbb dx$ cannot be $1$ (or $-1$, with the same argument). Consider the change of variable $y = -x$; this turns the integral into
$$\int \limits _{-\varepsilon} ^\varepsilon \mathrm{sgn}(x) \delta(x) \ \Bbb dx = \int \limits _\varepsilon ^{-\varepsilon} \mathrm{sgn} (-y) \delta(-y) \ (-\Bbb d y) = \int \limits _{-\varepsilon} ^\varepsilon \big( -\mathrm{sgn} (y) \big) \delta(y) \ \Bbb d y = -\int \limits _{-\varepsilon} ^\varepsilon \mathrm{sgn} (y) \delta(y) \ \Bbb d y ,$$
because $\delta$ is known to be even (this claim can be made to be rigorously correct) and $\mathrm{sgn}$ is clearly odd. This shows that, at best, $\lim _{\varepsilon \to 0} \int _{-\varepsilon} ^\varepsilon \mathrm{sgn}(x) \delta(x) \ \Bbb dx$ may be given the value $0$ - but this trick works because we have assumed that $\delta$ behaves like a function under changes of variable (again, this may be made rigorous but it requires some serious prerequisites).
The trouble begins if you replace $\delta$ by a function that is not odd - you don't know what to do anymore, then (try it, along the line sketched above). There are two major solutions out of this difficulty, both of them requiring the creation of a whole theory.
Solution 1: Work in the context of the theory of distributions: either view $\delta$ as a Schwartz distribution acting on compactly-supported smooth functions, or as a tempered distribution acting on Schwartz functions, or as a compactly-supported distribution acting on smooth functions. In all these three options, $\delta$ must act on some class of smooth functions, which $\mathrm{sgn}$ and the other function are not, so this framework is not suited for your problem.
Solution 2: Work in the context of measure theory. In this case, if $f \in L^1 \big( (-\varepsilon, \varepsilon), \delta \big)$, then
$$\int \limits _{(-\varepsilon, \varepsilon)} f \ \Bbb d \delta = f(0) .$$
The question is: are your two functions ($\mathrm{sgn}$ and the one with branches) $\delta$-integrable? Since every subset of $(-\varepsilon, \varepsilon)$ is $\delta$-measurable, it follows that every $\Bbb C$-valued function defined on $(-\varepsilon, \varepsilon)$ is $\delta$-measurable and, with the formula above, $\delta$-integrable.
We deduce that in this framework $\mathrm {sgn}$ is $\delta$-integrable and
$$\int \limits _{(-\varepsilon, \varepsilon)} \mathrm{sgn} \ \Bbb d \delta = \mathrm{sgn}(0) = 0 ,$$
but the other function is not $\delta$-integrable because it is not defined in $0$.
My answer above is formulated under the convention that $\mathrm{sgn} (x) = \begin{cases} -1, & x<0 \\ 0, & x=0 \\ 1, & x>0 .\end{cases}$
Notice that $\lim _{\epsilon \to 0}$ plays absolutely no role in the whole problem, it could be missing as well.