= Solution
On the attracting right branch, the leading fast balance is $w=1-v$, so $\dot w=v=1-w$. On the left branch it is $w=-v$, so $\dot w=-w$. The slow travel times are therefore
$$
T_R=\int_{w_L}^{w_R}\frac{dw}{1-w}=\log\frac{1+\alpha}{\beta},\qquad T_L=\int_{w_R}^{w_L}\frac{dw}{-w}=\log\frac{1-\beta}{-\alpha}.
$$
Both are positive because $w_R>w_L$. The fast jumps contribute vanishing time as $\epsilon\to0$, giving the <piecewise linear relaxation-oscillator period>
$$
\boxed{T\sim\log\frac{1+\alpha}{\beta}+\log\frac{1-\beta}{-\alpha}=\log\frac{(1+\alpha)(1-\beta)}{-\alpha\beta}.}
$$
For the illustrated $\alpha=-0.25$, $\beta=0.4$, the leading period is $1.5041$ in the given nondimensional time units.
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