Limit of function

We concentrate on limits of functions at the limit points of their domains.

Example. Consider

 f(x) = x for x ≠ 1.

with

 D(f) = ( -∞, 1 ) ∪ ( 1, ∞ ).

Even though the function is undefined at 1, the limit

 limx→1 f(x)

makes sense, because 1 is a limit point of D(f).

Example. Let

 f(x) = 1,
 D(f) = { 0 } ∪ [ 0, 2 ).

Here the limit

 limx→0 f(x)

does not make sense.

Example. Also, if

 D(f) = ( 1, ∞ ),

then the limit

 limx→∞ f(x)

does make sense.

Definition. Suppose

 f: ℝn → ℝ

and a a limit point of D(f). Then L ∈ ℝ is the limit of f as x → a if

 for any ε > 0 there is a δ > 0 such that 
 ‖ x - a ‖ < δ with x ∈ D(f) =>  | f(x) - L | < ε.

Note 1: there is always such an x - on the right - for any δ > 0.)

Note 2: if f: ℝⁿ → ℝ^m, then L ∈ ℝ^m and we replace in the definition the absolute value |.| with the norm ‖.‖ , as before:

 ‖ x - a ‖ < δ with x ∈ D(f) =>  | f(x) - L | < ε.

Note 3: Is the limit well-defined?

Theorem (Uniqueness of Limit). If L and M satisfy the definition, then L = M.

Proof'.' Let L ≠ M, L, M ∈ ℝ. Say

 L > M,

then let

 ε = ( L - M ) / 2.

Apply the definition with this ε. Then there is a δ satisfying the condition. Now choose x ∈ D(f) with

 ‖ x - a ‖ < δ.

Then it follows that

 | f(x) - L | < ε,
 | f(x) - M | < ε.

(Provide the details here!) Then

 | f(x) - L | + | f(x) - M | < 2ε = L - M.

Finish the proof. QED

Theorem. Let

 limx→a f(x) = L,
 limx→a g(x) = M.

Then

 (1) limx→a ( f(x) + g(x) ) = L + M,
 (2) limx→a ( f(x) ⋅ g(x) ) = L ⋅ M,
 (3) limx→a ( f(x) / g(x) ) = L / M, provided M ≠ 0.

Theorem (Limits Under Composition). For the two cases

 (1) f: ℝ → ℝn, g: ℝn → ℝ,
 (2) f: ℝn → ℝ, g: ℝ → ℝn

let

 limx→a f(x) = L,
 limt→L g(t) = M.

Then

 limx→a ( g ∘f )(x) = M.

Theorem. The function f is continuous at x = a iff

 limx→a f(x) = f(a).

The point a is a limit point of D(f).

Example. Let

 f( x, y ) = x + y continuous on D(f),
 D(f) = { ( x, 0 ): x ∈ ℝ } ⊂ ℝ2.

Consider points in D(f) that have a ball around them that lies entirely in D(f), interior point D(f). (This is unlike the first example.)

Definition. Given A ⊂ ℝⁿ, the point a ∈ A is called an interior point of A if there is an ε > 0 such that

 B( a, ε ) ⊂ A.

Let n = 1. Is a an interior point of A = { a }? No.

A = [ 0, 1 ]. What are the interior points? All but 0 and 1.

The set of all interior points of A is called the interior of A, and we write int A.

The interior

 int [ 0, 1 ] = ( 0, 1 ),

as the limit points of [ 0, 1 ] are [ 0, 1 ], so 0 and 1 are limit points but not interior points.

Is vice versa possible? No. Every interior point is a limit point (see Theorem).

Consider

 B( a, r ) = { x ∈ ℝn: | x - a | ≤ r }, a closed ball,

then

 int B( a, r ) = { x ∈ ℝn: | x - a | < r },

i.e. an open ball.