Discontinuity of Logarithmic Functions

The logarithmic function is an important type of function in calculus and can exhibit discontinuities in certain cases.

Understanding the discontinuity of logarithmic functions is essential for mastering the concept of discontinuity.

Basic Properties

The logarithmic function has the following basic properties:

  • Monotonicity: the logarithmic function is a monotone function
  • Continuity: continuous on its domain
  • Discontinuity: discontinuous at certain boundary points
  • Inverse relation: the logarithmic function and the exponential function are inverses of each other

Basic Logarithmic Functions

The Natural Logarithmic Function

Logarithmic Function

f(x)=ln⁡xf(x) = \ln x

Properties:

  • Domain: (0,+∞)(0, +\infty)
  • Continuous on its domain
  • The graph is a monotonically increasing logarithmic curve
  • Range: R\mathbb{R}
  • Discontinuous at x=0x = 0 (the function is undefined there)

The General Logarithmic Function

General Form of the Logarithmic Function

f(x)=log⁡axf(x) = \log_a x (a>0,a≠1a > 0, a \neq 1)

Properties:

  • Domain: (0,+∞)(0, +\infty)
  • Continuous on its domain
  • Monotonically increasing when a>1a > 1, monotonically decreasing when 0<a<10 < a < 1
  • Range: R\mathbb{R}
  • Discontinuous at x=0x = 0 (the function is undefined there)

Analysis of the Points of Discontinuity of Logarithmic Functions

Point of Discontinuity: x=0x = 0

Analysis:

  • The function is undefined at x=0x = 0
  • Right limit: lim⁡x→0+ln⁡x=−∞\lim_{x \to 0^+} \ln x = -\infty
  • This is a discontinuity of the second kind

Characteristics:

  • The function values tend to negative infinity
  • There is a vertical asymptote x=0x = 0
  • The limit does not exist (it tends to infinity)

Discontinuity of Composite Logarithmic Functions

Example 1: f(x)=ln⁡(x2−1)f(x) = \ln(x^2 - 1)

Analysis:

  • The inner function h(x)=x2−1h(x) = x^2 - 1 is continuous on R\mathbb{R}
  • The outer function g(x)=ln⁡xg(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  • We need x2−1>0x^2 - 1 > 0, that is, x<−1x < -1 or x>1x > 1
  • Therefore the interval of continuity is (−∞,−1)∪(1,+∞)(-\infty, -1) \cup (1, +\infty)
  • It is discontinuous at x=−1x = -1 and x=1x = 1

Analysis of the points of discontinuity:

  • Points of discontinuity: x=−1x = -1 and x=1x = 1
  • At these points, the inner function satisfies x2−1=0x^2 - 1 = 0
  • The logarithmic function is undefined at 0
  • These are boundary discontinuities

Example 2: f(x)=ln⁡(cos⁡x)f(x) = \ln(\cos x)

Analysis:

  • The inner function h(x)=cos⁡xh(x) = \cos x is continuous on R\mathbb{R}
  • The outer function g(x)=ln⁡xg(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  • We need cos⁡x>0\cos x > 0
  • Therefore the interval of continuity is all the values for which cos⁡x>0\cos x > 0 holds, i.e. all the xx values that satisfy it
  • It is discontinuous at every point where cos⁡x=0\cos x = 0

Analysis of the points of discontinuity:

  • Points of discontinuity: x=π2+kπx = \frac{\pi}{2} + k\pi (k∈Zk \in \mathbb{Z})
  • At these points, cos⁡x=0\cos x = 0
  • The logarithmic function is undefined at 0
  • These are periodic points of discontinuity

Example 3: f(x)=ln⁡(x2+1)f(x) = \ln(x^2 + 1)

Analysis:

  • The inner function h(x)=x2+1h(x) = x^2 + 1 is continuous on R\mathbb{R}
  • The outer function g(x)=ln⁡xg(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  • Since x2+1>0x^2 + 1 > 0 holds for every x∈Rx \in \mathbb{R}
  • Therefore f(x)=g(h(x))f(x) = g(h(x)) is continuous on R\mathbb{R}

Conclusion: this function is continuous everywhere on R\mathbb{R} and has no point of discontinuity.

Types of Point of Discontinuity

The points of discontinuity of logarithmic functions mainly fall into the following types:

  1. Discontinuities of the second kind: the limit tends to infinity
  2. Undefined points: the function is undefined at that point
  3. Boundary discontinuities: discontinuous at the boundary of the domain

Determining Continuity

  1. Basic logarithmic functions: continuous on (0,+∞)(0, +\infty)
  2. Composite logarithmic functions: the range of the inner function must be checked to lie within (0,+∞)(0, +\infty)
  3. Piecewise logarithmic functions: the left limit, the right limit and the function value at the breakpoint must be checked

Exercises

Exercise 1

Determine the continuity of f(x)=ln⁡(x2+1)f(x) = \ln(x^2 + 1) on R\mathbb{R}.

logarithmic functioncontinuityfunction continuity

Approach: Analyse the continuity of the composite function.

Steps:

  1. The inner function g(x)=x2+1g(x) = x^2 + 1 is continuous on R\mathbb{R}
  2. The outer function h(x)=ln⁡xh(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  3. Since x2+1>0x^2 + 1 > 0 holds for every x∈Rx \in \mathbb{R}
  4. Therefore f(x)=h(g(x))f(x) = h(g(x)) is continuous on R\mathbb{R}

Answer: The function is continuous on R\mathbb{R}.

Exercise 2

Determine the intervals of continuity of f(x)=ln⁡(x2−1)f(x) = \ln(x^2 - 1).

logarithmic functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the composite function.

Steps:

  1. The inner function g(x)=x2−1g(x) = x^2 - 1 is continuous on R\mathbb{R}
  2. The outer function h(x)=ln⁡xh(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  3. We need x2−1>0x^2 - 1 > 0, that is, x<−1x < -1 or x>1x > 1
  4. Therefore the interval of continuity is (−∞,−1)∪(1,+∞)(-\infty, -1) \cup (1, +\infty)

Answer: The interval of continuity is (−∞,−1)∪(1,+∞)(-\infty, -1) \cup (1, +\infty).

Exercise 3

Determine the intervals of continuity of f(x)=ln⁡(cos⁡x)f(x) = \ln(\cos x).

logarithmic functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the composite function.

Steps:

  1. The inner function g(x)=cos⁡xg(x) = \cos x is continuous on R\mathbb{R}
  2. The outer function h(x)=ln⁡xh(x) = \ln x is continuous on (0,+∞)(0, +\infty)
  3. We need cos⁡x>0\cos x > 0
  4. Therefore the interval of continuity is all the values for which cos⁡x>0\cos x > 0 holds, i.e. all the xx values that satisfy it

Answer: The interval of continuity is all the values for which cos⁡x>0\cos x > 0 holds, i.e. all the xx values that satisfy it.

Exercise 4

Determine the type of discontinuity of f(x)=ln⁡xf(x) = \ln x at x=0x = 0.

logarithmic functioncontinuityfunction continuity

Approach: Analyse the limiting behaviour of the function at that point.

Steps:

  1. The function is undefined at x=0x = 0
  2. Right limit: lim⁡x→0+ln⁡x=−∞\lim_{x \to 0^+} \ln x = -\infty
  3. The limit tends to negative infinity, so there is no finite limit
  4. This is a discontinuity of the second kind

Answer: x=0x = 0 is a discontinuity of the second kind.

Exercise 5

Determine the continuity of f(x)=ln⁡∣x∣f(x) = \ln|x| at x=0x = 0.

logarithmic functioncontinuityfunction continuity

Approach: Analyse whether the function is defined at that point.

Steps:

  1. When x=0x = 0, ∣x∣=0|x| = 0
  2. ln⁡0\ln 0 is undefined
  3. Therefore the function is undefined at x=0x = 0
  4. The function is discontinuous at x=0x = 0

Answer: The function is discontinuous at x=0x = 0, because the function is undefined at that point.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
R\mathbb{R}Mathematical symbolBlackboard bold R (Real numbers)The set of all real numbers
ln⁡\lnMathematical symbolNatural logarithmThe logarithm with base ee
Z\mathbb{Z}Mathematical symbolBlackboard bold Z (Integers)The set of all integers

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