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:

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