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.

I doubt that the method with the limit is always possible, but I am also looking forward to an answer that explains if it is applicable at all or if it is applicable under certain conditions. 
This is the reason why this is not an answer for the applicability of the limit for the stability analysis of this system. I will rather show a direct method to investigate the stability of the nonlinear system.
First, note that $x=0$ is an equilibrium point of the nonlinear system. Then integrate the equation exploiting that $\text{sgn}(x)$ can be treated as a constant.
$x(t)=-\text{sgn}(x)t+c$
Edit: In order to motivate this solution you can look at three cases $x>0$, $x=0$ and $x<0$. For the first case, we obtain $$\dot{x}=-1 \implies x(t) = -t+ c_1.$$ The case $x=0$ leads to $$\dot{x}=0 \implies x(t) = c_2.$$
The last case results in
$$\dot{x} = +1 \implies x(t) = t + c_3$$
Combining all cases gives the previous expression.
If $x(t=0)=x_0$ then we can look at the two cases $x_0>0$ and $x_0<0$. Can you continue?
