MTH6131 Chap.3 Continuity and the Value Theorems
Continuity and the Value Theorems
Continuity is decided point by point
The lecture warns against the everyday picture of continuity as a graph drawn without lifting the pen. Continuity is defined at points of the domain: f is continuous at c when for every positive ε there is a positive δ such that inputs within δ of c give outputs within ε of f(c).
In practice this is checked through three conditions at the point: the value is defined, the limit exists, and the limit equals the value. A function passing all three at every point of its domain is continuous there.
Sorting the ways to fail
The exercises ask not just whether a function is continuous but how it fails when it is not.
If the limit exists but the value is missing or wrong, the discontinuity is removable: define or redefine one value and it disappears. Cancelling a common factor in a rational function is the usual evidence. If the two one-sided limits are finite but different, the discontinuity is a jump, and no single value repairs it. If one side grows without bound, the discontinuity is infinite.
Absolute values and piecewise definitions are where the one-sided limits need computing separately.
What continuity allows
Sums, differences and products of continuous functions are continuous, and so are quotients where the denominator is not zero. This is why polynomials are continuous everywhere and rational functions everywhere except at zeros of the denominator.
On a closed interval, continuity guarantees two further things. The Extreme Value Theorem promises a least and a greatest value, and the Intermediate Value Theorem promises that every value between f(a) and f(b) is taken somewhere in between. The second is the basis of locating roots by a change of sign.
Why the hypotheses matter
Both theorems collapse without their conditions.
A function on an interval that is not closed may have no maximum, and a function with a jump can skip a value entirely. Checking the hypotheses before quoting a theorem is a habit this chapter starts and every later chapter relies on.
What this chapter covers
- 01
Continuity defined at a point, not by a picture
- 02
The three-condition test
- 03
Removable discontinuities and cancelled factors
- 04
Jump and infinite discontinuities
- 05
Absolute values and piecewise joins
- 06
Arithmetic of continuous functions
- 07
The Extreme Value Theorem
- 08
The Intermediate Value Theorem and root location
Worked example · free
Two discontinuities of different kinds
- 1Find where the denominator is zero.
- 3Factor and cancel any common factor, then take the limit at each point.
- 2Classify each point by the result.
Key terms
- Jump Discontinuity
- A point where both one-sided limits exist and are finite but differ, so that no choice of value can make the function continuous there.
- Infinite Discontinuity
- A point where at least one one-sided limit is unbounded, typically at a zero of a denominator that does not cancel.
- Closed Interval
- An interval that includes both of its endpoints, the setting in which the Extreme Value and Intermediate Value Theorems apply.
- Extreme Value Theorem
- The result that a function continuous on a closed interval attains both a least and a greatest value on that interval.
- Intermediate Value Theorem
- The result that a function continuous on a closed interval takes every value between its values at the two endpoints.
Continuity and the Value Theorems FAQ
Is a function with a hole in its graph discontinuous?
It depends on whether the hole is in the domain. If the formula is undefined at the point, the function is simply not defined there; the useful question is whether it can be extended continuously, which it can when the limit exists. If the function is defined at the point with a different value, it is discontinuous there, and redefining that one value removes the discontinuity.
How do I prove an equation has a root without solving it?
Rewrite it as f(x) = 0 with f continuous on a closed interval, and find two endpoints where f has opposite signs. The Intermediate Value Theorem then guarantees a root in between. Halving the interval and keeping the half with the sign change narrows it down as far as you need.
Why must the interval be closed for the Extreme Value Theorem?
Because on an open or half-open interval the function can approach a largest or smallest value without reaching it. The function 1/x on the interval from 0 to 1, excluding 0, grows without bound near zero and so has no greatest value, even though it is continuous at every point where it is defined.
Exam move
For every function in the exercises, write the three conditions as three separate lines at each suspicious point, even when the answer seems obvious. The habit pays off on piecewise and absolute value formulas, where the one-sided limits use different pieces and a quick glance gives the wrong verdict.
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