Prove that $\displaystyle \lim _{x \rightarrow a } f(x) = \alpha $ if and only if $\displaystyle \limsup _{x \rightarrow a} f(x) = \liminf _{x \rightarrow a} f(x) = \alpha $.

Proof so far:

Suppose that $\displaystyle \limsup _{x \rightarrow a} f(x) = \liminf _{x \rightarrow a} f(x) = \alpha $, then for $\displaystyle \delta > 0 $, we have:

$\displaystyle \inf _{ \delta > 0} \sup _{0< \mid x - \alpha \mid < \delta } f(x) = \sup _{ \delta > 0} \inf _{0< \mid x - \alpha \mid < \delta } f(x) = \alpha $

Implies that $\displaystyle \forall \epsilon > 0 \exists \delta _1 , \delta _2 > 0 $ such that:

$\displaystyle \sup _{0< \mid x - \alpha \mid < \delta } f(x) = \alpha + \epsilon $ and $\displaystyle \liminf _{x \rightarrow a} f(x) = \alpha - \epsilon $

Now, define $\displaystyle \delta = max \{ \delta _1 , \delta _2 \} $, we have:

$\displaystyle \alpha - \epsilon < f(x) < \alpha + \epsilon $

$\displaystyle \mid f(x) - \alpha \mid < \epsilon $ when $\displaystyle \mid x - \alpha \mid < \delta $

So then $\displaystyle \lim _{x \rightarrow a} f(x) = \alpha $

Am I okay so far? I know the idea should be correct, but I worry about the way I write them and the technical details may not be clear.