# Limit, Limit Superior, and Limit Inferior of a function

• Sep 3rd 2009, 12:28 PM
Limit, Limit Superior, and Limit Inferior of a function
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.
• Sep 3rd 2009, 12:41 PM
Matt Westwood
Here's an exposition of this theorem:

Convergence of Limsup and Liminf - ProofWiki

which simplifies the notation slightly. This time of night all those subscripts and epsilons and deltas are making my head swim, so I'm utterly unable to compare yours with this and make sense of it ... it's been a long day.
• Sep 3rd 2009, 05:05 PM