Sliding mode stability Given the scalar system
$$
\dot{x} = -\text{sgn}(x) \tag{1}
$$
with
$$
\text{sgn}(x) = \begin{cases}
-1 & x < 0 \\
0 & x = 0 \\
1 & x > 0 \,.
\end{cases}
$$
What is an easy method to check for stability? And: Say I use quasi sliding mode with
$$
\dot{x} = -\tanh(a x) \,. \tag{2}
$$
If it is shown that $(2)$ is stable for all $a > 0$, does that also show stability of $(1)$ because
$$
\lim_{a \rightarrow \infty} \tanh(a x) = \text{sgn}(x) \,?
$$
Edit: For the general question, it is required that both original and aproximation function have exactly the same equilibria, i.e. that the approximation doesn't add any new equilibira.

The answer to the second question is generally NO.  Consider a family $f_a(x) = x (x - \frac{1}{a}) (x + \frac{1}{a})$, $a > 0$.  One has $\lim\limits_{a \to \infty} f_a(x) = x^3$, uniformly for $x$ in compact subsets of $\mathbb{R}$.  The equilibrium $0$ is (even asymptotically) stable for any $\dot{x} = f_a(x)$, whereas it is unstable for the limiting equation $\dot{x} = x^3$.
