Limit Method

Use the properties of limits to determine continuity.

Determination Principles

  • If lim⁡x→x0f(x)=f(x0)\lim_{x \to x_0} f(x) = f(x_0), then the function is continuous at x0x_0
  • If the limit does not exist, or the limit is not equal to the function value, then the function is discontinuous at that point

Applied Examples

Example 1: determine the continuity of the function f(x)=sin⁡xf(x) = \sin x at x=0x = 0

Solution:

  1. f(0)=sin⁡0=0f(0) = \sin 0 = 0
  2. lim⁡x→0sin⁡x=0\lim_{x \to 0} \sin x = 0
  3. lim⁡x→0f(x)=f(0)\lim_{x \to 0} f(x) = f(0)
  4. Conclusion: the function is continuous at x=0x = 0

Example 2: determine the continuity of the function f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1} at x=1x = 1

Solution:

  1. The function is undefined at x=1x = 1
  2. lim⁡x→1x2−1x−1=lim⁡x→1(x+1)=2\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2
  3. The limit exists but the function value is undefined
  4. Conclusion: the function is discontinuous at x=1x = 1

Exercises

Exercise 1

Determine the continuity of f(x)={sin⁡xx,x≠01,x=0f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} at x=0x = 0.

determination of continuitylimit method

Approach: Apply the limit method.

Steps:

  1. The function is defined at x=0x = 0: f(0)=1f(0) = 1
  2. Compute the limit: lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
  3. Compare: lim⁡x→0f(x)=f(0)=1\lim_{x \to 0} f(x) = f(0) = 1

Answer: The function is continuous at x=0x = 0.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
x0x_0Mathematical symbolx-sub-zeroThe point at which continuity is decided from the limit
f(x)f(x)Mathematical symbolf of xThe function whose continuity is under examination
f(x0)f(x_0)Mathematical symbolf of x-sub-zeroThe value the function takes at x0x_0
lim⁡x→x0f(x)\lim_{x \to x_0} f(x)Mathematical symbollimit of f of xThe limit whose value decides continuity at x0x_0
sin⁡x\sin xMathematical symbolsine of xThe function used in the first worked example
x2−1x−1\frac{x^2 - 1}{x - 1}Mathematical symbolrational expressionThe quotient whose limit is 2 at the undefined point x=1x = 1
sin⁡xx\frac{\sin x}{x}Mathematical symbolsine x over xThe quotient whose limit is 1 at the origin, used in the exercise
≠\neqMathematical symbolnot equal toThe relation that separates a continuous point from a broken one

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