How can I calculate the limit value of this floor function?

[stiff pawn]

$$\lim{n \to \infty} \lfloor n/2\rfloor$$
Sorry for the shitty LaTeX

[slim shard]

DVD

[alpine scarab]

$\lim_{n\to\infty}\floor{\frac{n}2}$

[slim shard]

A Lonely Bean

[alpine scarab]

Doesn't the limit seem to diverge?

[stiff pawn]

I don't understand, Where is this diverge?

[alpine scarab]

It goes to infinity as you look to the right

[stiff pawn]

Yes but how can I proof it

[alpine scarab]

There is a theorem in analysis saying that a limit of f as x -> a exists iff every sequence x_n such that x_n -> x, the limit of f(x_n) exists n -> infinity and is equal to the same constant for all sequences

So, let's do a proof by contradiction

Assume $\lim_{x\to\infty}\floor{\frac{x}2}$ exists

[slim shard]

A Lonely Bean

[alpine scarab]

Now pick $x_n = 2n \forall n \in \mathbb{Z}^+$, according to the theorem the limit of $\floor{\frac{x_n}2}$ should exist because $x_n$ tends to infinity like $x$ does

[slim shard]

A Lonely Bean

[stiff pawn]

Ah yes

[alpine scarab]

But what you see is that $\lim_{n\to\infty}\floor{\frac{x_n}2}$ diverges

[slim shard]

A Lonely Bean

[alpine scarab]

Since $\floor{\frac{x_n}2}=n$

[slim shard]

A Lonely Bean

[alpine scarab]

Referring to this theorem is really helpful when you are dealing with the floor function or a non-constant periodic function

[stiff pawn]

Ok thank, I understood 🙂

[cinder gorge]

.solved