# Limits and continuous functions

## Limits

### Limits of real functions

#### Limit operator

For a function $$f(x)$$,

$$\lim_{x\rightarrow a}f(x)=L$$

We can say that $$L$$ is the limit if:

$$\forall \epsilon >0 \exists \delta >0 \forall x[0<|x-p|<\delta \rightarrow |f(x)-L|<\epsilon]$$

### Limit superior and limit inferior

If a sequence does not converge, but stays between two points, then lim sup is upper bound, lim inf is lower bound.

## Continuous functions

### Continous functions

A function is continuous if:

$$\lim_{x\rightarrow c} f(x)=f(c)$$

For example a function $$\dfrac{1}{x}$$ is not continuous as the limit towards $$0$$ is negative infinity. A function like $$y=x$$ is continous.

More strictly, for any $$\epsilon >0$$ there exists

$$\delta >0$$

$$c-\delta < x< c +\delta$$

Such that

$$f(c)-\epsilon < f(x) < f(c)+\epsilon$$

This means that our function is continuous at our limit $$c$$, if for any tiny range around $$f(c)$$, that is $$f(c)-\epsilon$$ and $$f(c)+\epsilon$$, there is a range around $$c$$, that is $$c-\delta$$ and $$c+ \delta$$ such that all the value of $$f(x)$$ at all of these points is within the other range.

#### Limits

Why can’t we use rationals for analysis?

If discontinous at not rational number, it can still be continous for all rationals.

Eg $$f(x)=-1$$ unless $$x^2>2$$, where $$f(x)=1$$.

Continous for all rationals, because rationals dense in reals.

But can’t be differentiated.

### Reals or rationals for analysis

Why can’t we use rationals for analysis?

If discontinous at not rational number, it can still be continous for all rationals.

eg $$f(x)=-1$$ unless $$x^2>2$$, where $$f(x)=1$$.

Continous for all rationals, because rationals dense in reals

But can’t be differentiated

### Boundedness theorem

If $$f(x)$$ is closed and continuous in $$[a,b]$$ then $$f(x)$$ is bounded by $$m$$ and $$M$$. That is:

$$\exists m \in \mathbb{R} \exists M\in \mathbb{R}\forall x\in [a,b](m<f(x)<M$$

### Intermediate value theorem

Take a real function $$f(x)$$ on closed interval $$[a,b]$$, continuous on $$[a,b,]$$.

IVT says that for all numbers $$u$$ between $$f(a)$$ and $$f(b)$$, there is a corresponding value $$c$$ in $$[a,b]$$ such that $$f(c)=u$$.

That is:

$$\forall u \in [min(f(a),f(b)),max(f(a),f(b))] \exists c \in [a,b] (f(c)=u)$$

### Extreme value theorem

We can expand the boundedness theorem such that $$m$$ and $$M$$ are functions of $$f(x)$$ in the bound $$[a,b]$$. That is:

$$\exists m \in \mathbb{R} \exists M\in \mathbb{R}\forall x\in [a,b](m<f(x)<M)$$