Continuity Notes
Continuity: Part 1 (Section 1.8, Page 81)
Intuitive Definition of Continuity
A function is continuous if its graph can be drawn without lifting the pencil from the paper.
This is an intuitive, not mathematical, definition.
Mathematical Definition of Continuity (Definition 1)
A function f is continuous at a number a if \lim_{x \to a} f(x) = f(a).
If a function is not continuous at a, it is discontinuous at a.
Three Conditions for Continuity
Based on the mathematical definition \lim_{x \to a} f(x) = f(a), a function f must satisfy three conditions to be continuous at a number a:
**Existence of f(a): **
f(a) must be defined (i.e., it must exist).
If f(a) does not exist, the function is not continuous at a.
**Existence of the Limit \lim_{x \to a} f(x): **
The limit of f(x) as x approaches a must exist.
For the limit to exist, the left-hand limit must equal the right-hand limit: \lim{x \to a^-} f(x) = \lim{x \to a^+} f(x).
**Equality of the Limit and Function Value: **
If the limit exists, it must be equal to the function value at a: \lim_{x \to a} f(x) = f(a).
Continuity from the Left and Right
Continuity from the Right: A function f is continuous from the right at a if \lim_{x \to a^+} f(x) = f(a).
Continuity from the Left: A function f is continuous from the left at a if \lim_{x \to a^-} f(x) = f(a).
Example 1: Determining Discontinuities from a Graph
Given the graph of a function F, determine the numbers at which F is discontinuous and explain why.
For each number where F is discontinuous, determine if F is discontinuous from the right, from the left, or neither.
Part A: Identifying Discontinuities
Examine the graph to find points where the three conditions for continuity are not met.
At A = 4 (or x = 4):
F(4) does not exist (open circle on the graph).
At A = -2 (or x = -2):
\lim{x \to -2} F(x) does not exist because \lim{x \to -2^-} F(x) \neq \lim_{x \to -2^+} F(x).
The limit from the left and the limit from the right approach different points.
At A = 2 (or x = 2):
\lim{x \to 2} F(x) does not exist because \lim{x \to 2^-} F(x) \neq \lim_{x \to 2^+} F(x).
At A = -4 (or x = -4):
\lim{x \to -4} F(x) does not exist because \lim{x \to 4^-} F(x) \neq \lim_{x \to 4^+} F(x).
\lim_{x \to 4^-} F(x) tends to infinity.
Part B: Determining Continuity from the Left or Right
Check the conditions for continuity from the left and right at each point of discontinuity.
At a = -4:
F(-4) does not exist (open circle).
Not continuous from the left, nor from the right.
At a = -2:
\lim_{x \to -2^-} F(x) = F(-2), because there is a shaded dot.
Continuous from the left.
Not continuous from the right (open circle).
At a = 2:
Not continuous from the left, there is no limit because f(2) is not defined.
Continuous from the right, there is a shaded dot, so f(2) is defined.
At a = 4:
Not continuous from the left, there is an asymptote at x=4, f(4) does not exist.
Continuous from the right, there is a shaded dot, so f(4) is defined.
Jump Discontinuities
Discontinuities at x = 2 and x = -2 are jump discontinuities.
Continuity on a Closed Interval (Definition 3)
If a function F is defined on a closed interval [A, B], then F is continuous on this interval if:
F is continuous from the right at A (
F is continuous from the left at B (
F is continuous on the open interval (A, B).
Example 2: Checking Continuity on Intervals
Interval [-1, 1]:
A = -1, B = 1
F is continuous from the right of A = -1 (no discontinuity).
F is continuous from the left of B = 1 (no discontinuity).
F is continuous on the open interval (-1, 1).
Yes, F is continuous on this interval.
Interval [-2, 0]:
A = -2, B = 0
F is not continuous from the right of A = -2 because F(-2) is not defined.
Therefore, F is not continuous on the interval [-2, 0] even without checking other conditions.
Interval [4, 6]:
A = 4, B = 6
F is continuous from the right of A = 4 (shaded point indicates definition).
F is continuous from the left of B = 6 (no discontinuity).
F is continuous on the open interval (4, 6).
Yes, F is continuous on this interval.
Theorem: Properties of Continuous Functions
If f and g are continuous at a, and c is a constant, then the following functions are also continuous at a:
f + g (sum)
f - g (difference)
cf (constant multiple)
fg (product)
\frac{f}{g} (quotient), provided g(a) \neq 0
These properties can be proven using the limit laws from Section 1.6.
Proof of Property 4 (Product of Continuous Functions)
Objective: Show that if f and g are continuous at a, then fg is continuous at a, i.e., \lim_{x \to a} (f(x)g(x)) = f(a)g(a).
Direct Proof:
Assume f and g are continuous at a.
This means \lim{x \to a} f(x) = f(a) and \lim{x \to a} g(x) = g(a) by the definition of continuity.
Consider \lim_{x \to a} (f(x)g(x)).
By the product law of limits (Section 1.6), \lim{x \to a} (f(x)g(x)) = \lim{x \to a} f(x) \cdot \lim_{x \to a} g(x).
Since f and g are continuous, this becomes f(a)g(a).
Therefore, \lim_{x \to a} (f(x)g(x)) = f(a)g(a), proving that fg is continuous at a.
Theorem: Continuity of Polynomials and Rational Functions
Any polynomial is continuous everywhere (on its domain).
Any rational function is continuous wherever it is defined (on its domain).
Refer to page 86 for proofs.
Example 3: Applying the Theorem
3.1 Finding the Limit of a Polynomial
Find \lim_{x \to 1} (x^5 + 2x^3 - 3x^2 + x).
Since x^5 + 2x^3 - 3x^2 + x is a polynomial of degree 5, it is continuous everywhere.
By the definition of continuity, \lim_{x \to a} f(x) = f(a).
Therefore, \lim_{x \to 1} (x^5 + 2x^3 - 3x^2 + x) = (1)^5 + 2(1)^3 - 3(1)^2 + (1) = 1 + 2 - 3 + 1 = 1.
3.2 Finding the Limit of a Rational Function
Find \lim_{x \to 3} \frac{x^2 - 2x}{1 - x}.
Since \frac{x^2 - 2x}{1 - x} is a rational function, it is continuous on its domain.
The domain is all x such that x \neq 1.
Since 3 is in the domain, we can evaluate the limit by direct substitution.
\lim_{x \to 3} \frac{x^2 - 2x}{1 - x} = \frac{(3)^2 - 2(3)}{1 - 3} = \frac{9 - 6}{-2} = \frac{3}{-2} = -\frac{3}{2}.