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

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

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

Proof so far:

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

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

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

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

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

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

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

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.

2. 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.

3. The proof on the website shows this fact with sequence. But I'm working on the limit of functions, in which I understand the idea is the same; however, I'm worry about my notation and the deltas, I'm not sure if they are all in the right places.

Thanks!

### limit inferior of a function

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