Suppose $\lim \limits_{x\to x_0} f(x) = L$, i.e., suppose:
$\forall \epsilon>0, \exists \delta>0, \forall x$ such that $0<|x-x_0| < \delta \implies |f(x) - L| < \epsilon$.
Let $\epsilon$ be given. Then, $\exists \delta>0$ such that $0<|x-x_0|<\epsilon \implies |f(x)-L| < \epsilon$.
Let $\alpha_1 = \frac{\epsilon}{5}$. Then, $\exists \alpha_2$ s.t. $0<|x-x_0|<\delta \implies |f(x)-L|<a_1=\frac{\epsilon}{5}$.