Continuity

at a point

A function f:𝒟→ℛ is continuous at ω if

  1. f is defined at ω i.e., ω∈𝒟;
  2. limx→ω f(x) exists;
  3. limx→ω f(x) =f(ω) .

One sided

Right continuous
limx→ω+ f(x) = f(ω)
Left continuous
limx→ω- f(x) = f(ω)

on interval

A function is continuous on an interval ℐ if it is continuous at every point in ℐ

Continuity properties

  • 𝒶 f(x) ± 𝒷 g(x)
  • f(x) g(x)
  • f(x) g(x) when defined
  • f(x) n for n∈ℤ+
  • f(x) n , n∈ℤ+ when defined

Polynomial and rational functions are continuous on their domain.

Trigonometric functions are continuous on their domains.

Inverse functions of continuous functions are continuous on their domain.

Discontinuities

removable
limx→ω f(x) ∈ℝ
jump

limx→ω+ f(x) ∈ℝ, limx→ω- f(x) ∈ℝ and limx→ω+ f(x) ≠ limx→ω- f(x)

Functions with removable discontinuities have continuous extensions.

Discontinuities

infinite
lim x→ω+ f(x) =±∞ or limx→ω- f(x) =±∞
oscillating
lim x→ω f(x) approaches more than one value

Explanations and Examples

Find the interval of continuity

  • f1 (x) = x-1 x2 +2⁢x
  • f2 (x) = x2 +2⁢x -5
  • f3 (x) = sin ( x2 -4 )

Evaluate the limits

  • limx→ π2 cos ( 2⁢x +sin ( 3⁢π 2 +x ) )
  • limx→0 x+1 ⁢ ℯtanx

Show that p(x) =x3 +3⁢x2 +x -1 has a root between zero and one.

Let f(x) = |x| x , then f(-3) =1 and f(2) =1, does that the Intermediate Value Theorem imply there is a root between negative three and two?

Implications

Composition of continuous functions

If f is continuous at ω and g is continuous at f(ω) then f∘g is continuous at ω

If lim x→ω f(x) =𝓁 and g(x) is continuous at 𝓁 then

limx→ω g (f(x) ) = g ( limx→ω f(x) )

Intermediate value theorem

Let f(x) be continuous on an interval [a,b] then.

∀ν∈ ( 𝓂,ℳ ) , ∃ ω ∈ (a,b) : f(ω)=ν

where

  • 𝓂 = min ( f(a) , f(b) ) , and
  • ℳ = max ( f(a) , f(b) )