Mathematics

Continuity

Study of the continuity of a function at a point and classification of discontinuities.

Continuity at a point: A function is continuous at a point x=ax=a if the following conditions are met:
  • The function is defined at aa, i.e., ∃f(a)\exists f(a)
  • The limit exists and is finite as xx approaches aa, which implies that the one-sided limits match:
lim⁡x→a−f(x)=lim⁡x→a+f(x)=k\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = k
  • The limit equals the function value, that is:
lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)
Continuity of a function: A function is said to be continuous if it is continuous at every point in its domain.
To analyze the continuity of a function, we follow these steps:
  1. Determine the possible points of discontinuity: based on these considerations:
    1. For piecewise functions, analyze the points where the function changes its definition.
    2. For rational functions, analyze the points that make the denominator zero.
    3. In general, analyze the endpoints of the domain intervals and points where the function is undefined.
  2. For each point x=ax=a of possible discontinuity, we will calculate:
lim⁡x→a−f(x);lim⁡x→a+f(x);f(a)\lim_{x\to a^-} f(x) \quad ; \quad \lim_{x\to a^+} f(x) \quad ; \quad f(a)
  1. The function is continuous at x=ax=a if:
lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)
  1. If the function is not continuous at x=ax=a, we will classify the discontinuity according to the adjacent table.
Example. Analyze the continuity of the function:
f(x)={1x−1x<02x−20≤xf(x) = \begin{cases} \frac{1}{x-1} & x < 0 \\ \frac{2}{x-2} & 0 \le x \end{cases}
1. We will study continuity at x=0x=0 (change of definition) and at x=2x=2 (makes the denominator of the 2nd function zero). It is not necessary to check x=1x=1 because, although it makes the denominator of the 1st function zero, it does not belong to its domain of definition.

2. Calculate the limits and the function value at each point:
Analyzing continuity at x=0x=0
lim⁡x→0−1x−1=−1\lim_{x\to 0^-} \frac{1}{x-1} = -1
lim⁡x→0+2x−2=−1\lim_{x\to 0^+} \frac{2}{x-2} = -1
f(0)=−1f(0) = -1
Analyzing continuity at x=2x=2
lim⁡x→2−2x−2=−∞\lim_{x\to 2^-} \frac{2}{x-2} = -\infty
lim⁡x→2+2x−2=+∞\lim_{x\to 2^+} \frac{2}{x-2} = +\infty
f(2)=∄f(2) = \nexists
3. The function is continuous at x=0x=0.
4. The function has an asymptotic (infinite) discontinuity at x=2x=2.
Removable Discontinuity
The one-sided limits exist, are finite, and are equal.
  • The function value f(a)f(a) does not exist:
Removable discontinuity without function value
  • The function value f(a)f(a) exists but does not equal the limits:
Removable discontinuity with function value
Non-Removable or Essential Discontinuity

Jump and Infinite Discontinuities (First kind):
  • Jump Discontinuity: the one-sided limits exist and are finite but are different:
Jump discontinuity 1
  • Infinite Jump Discontinuity: one of the one-sided limits is finite and the other is infinite:
Infinite jump discontinuity 1
  • Asymptotic (Infinite) Discontinuity: both one-sided limits are infinite:
Asymptotic discontinuity 1
Essential Discontinuity (Second kind): At least one of the one-sided limits does not exist:
Essential discontinuity 1