Limit of a Function
Let S ⊆ R n and f : S → R m . If a is a cluster point of S then a point v ∈ R m is the limit of f at a if for every ϵ ∈ R + there exists an r ∈ R + such that for any x ∈ S we have 0 < ‖ x − a ‖ < r ⟹ ‖ f ( x ) − v ‖ < ϵ and we write lim x → a f ( x ) = v
The reason we want a cluster point is that if it was not a cluster point, and for example if S = { 1 , 2 , 3 } and we're looking at the point 1 , f is the identity function, and then we claim that lim x → 1 f ( x ) = 1 then for any epsilon we can choose r small enough so that no point in the domain other than the point 1 is in the ball, then clearly this holds true, but that's not what we wanted when we designed the idea of a limit, we want to say that points of the function are all getting close to 1, therefore without the added assumption of cluter point, the definition fails the very design we wanted it to have.
Continuity of a Function at a Point
Suppose that S ⊆ R n and let f be a function f : S → R m we say that f is continuous at a ∈ S if for every ϵ ∈ R + there is some r ∈ R + such that for all x ∈ S we have ‖ x − a ‖ < r ⟹ ‖ f ( x ) − f ( a ) ‖ < ϵ
Continuous Function
We say that a function f is continuous if it is continuous for every x ∈ dom ⁡ ( f )

The above definition looks very much like the limit of a function, but note the difference, since in the limit we force that 0 < ‖ x − a ‖ this means that we don't have a positive implication for when x = a , in fact in such a case the implication is trivial which means that no information is gained in that case, whereas the definition of continuity of a function at a point does not have such a case, and therefore allows for knowledge gain when x = a and in such a case we must also have that ‖ f ( x ) − f ( a ) ‖ < ϵ , this implies that our function is actually well behaved at that point.

Discontinuity of a Function at a Point
We say that a function is discontinuous at a point if if it is not continuous there
The Characteristic Function is Discontinuous on the Boundary
Let A ⊆ R n then χ A is discontinuous for every point in ∂ A
Lipschitz
A function from S ⊆ R n into R m is said to be Lipschitz if there exists some C ∈ R + such that for all x , y ∈ S ‖ f ( x ) − f ( y ) ‖ ≤ C ‖ x − y ‖ the Lipschitz constant of f is the smallest C for which the condition holds.
Every Lipschitz Function is Continuous
As per title.
Lipschitz Condition of Order Alpha
Suppose that f : [ a , b ] → R and α ∈ R + then if there is some M ∈ R + such that for any x , y ∈ [ a , b ] we have | f ( x ) − f ( y ) | ≤ M | x − y | α Let lip ⁡ ( α ) denote the set of all functions satisfying a lipschitz condition of order α
Base and Exponent in the Range 0 to 1 Inequality
Suppose that a , b ∈ ( 0 , 1 ) then b < b a
x to the Alpha is in Lip Alpha
For α ∈ ( 0 , 1 ) show that x α ∈ lip ⁡ ( α ) where x α : R + → R +
Let x , y ∈ R + our goal is to find some M ∈ R + such that | x α − y α | ≤ M | x − y | α if x = y this is trivially true because 0 ≤ 0 so without loss of generality we will assume that x > y so that dividing by x α on both sides we obtain | 1 − ( y x ) α | ≤ M | 1 − y x | α allowing x α to enter into the absolute value because it is positive, and noticing that 1 > y x , set λ := y x therefore we're trying to prove (also removing some redudant absolute values): 1 − λ α ≤ M ( 1 − λ ) α so we have to prove that 1 ≤ M ( 1 − λ ) α + λ α by choosing a good value of M , but we clearly have 1 ≤ ( 1 − λ ) + λ then since ( 1 − λ ) , λ ∈ ( 0 , 1 ) and α ∈ ( 0 , 1 ) we know that ( 1 − λ ) α > ( 1 − λ ) and that λ α > λ therefore we have 1 ≤ ( 1 − λ ) + λ ≤ ( 1 − λ ) α + λ α therefore a selection of M = 1 proves what we have to show.