Chapter 05

Continuity: Unbroken Curves, Opening the Door to Derivatives

Continuity at a point means the limit exists and equals the function value there. It is the basis for differentiability and for understanding activation and loss functions in deep learning.

Math diagram by chapter

Select a chapter to see its diagram below. View the flow of basic math at a glance.

Left: continuous — the curve passes through $(a, f(a))$ with no gap or jump. Right: discontinuous — a hole or jump at that point.

Continuous

lim = f(a)

Discontinuous

f(a) missing or lim \neq f(a)

Continuity means $\lim_{x \to a} f(x) = f(a)$ . On the graph, the curve does not break at that point.

How to read

Left graph: $y = x^2$ is continuous at $x = 2$ (curve passes through (2, 4) with no break).
Right graph: if the function value is missing or differs from the limit at $x = a$ , it is discontinuous (hole or jump).

What is continuity?

Continuity is often described intuitively as a graph you can draw in one stroke without lifting the pencil. Mathematically, we need a precise condition: the limit as

x

approaches

a

must exist and equal the function value

f(a)

a

. We write

\lim_{x \to a} f(x) = f(a)

Three-step checklist for continuity at a point

x=a

(1) Function value exists:

f(a)

must be defined (no hole).

(2) Limit exists:

\lim_{x \to a} f(x)

must exist. The left limit (as

x

approaches

a

from below) and right limit (from above) must be equal; if one is 0 and the other 1 (like a step), the limit does not exist and the function is discontinuous.

(3) They match: the limit and

f(a)

must be equal (the path exists but the bridge must not be in the wrong place).

Discontinuity makes prediction unreliable—e.g. a stock at 100 yesterday and 0 today (jump), or missing data (hole). Mathematically, continuity is the promise that if input $x$ changes only a little ( $\delta$ ), output $f(x)$ also changes only a little ( $\varepsilon$ )—i.e. stability.

It is the prerequisite for differentiation. Differentiation is the slope of the tangent; if the graph is broken, there is no well-defined slope. So we need continuity before we can hope for differentiability. (Note: continuous does not always imply differentiable—a sharp corner is continuous but not differentiable.)

Robustness (avoiding butterfly effects): The AI model must be continuous so that a small amount of noise in the input does not cause a wildly wrong output. If a self-driving car misreads "stop" as "accelerate" because of a tiny scratch on a sign, that would mean the model behaved discontinuously—and that is very dangerous.

It is central to activation function design. ReLU, Sigmoid, Tanh and the like are all continuous, so information flows through the network without breaking. The loss function must be a smooth continuous surface so we can roll the ball (parameters) downhill to find the lowest point (the best answer)—gradient descent.

To check continuity at a point, verify: does $\lim_{x \to a} f(x)$ exist?, is $f(a)$ defined?, and are they equal?

Checklist:

①

f(a)

exists

②

\lim_{x \to a} f(x)

exists

③ limit

= f(a)

. If any fails, the function is discontinuous at that point.

Example problems and solutions are in the table below.

Problem $f(x) = x^2$
Solution $f(2) = 4$

Problem $g(x) = \frac{1}{x}$
Solution $g(0)$

Problem $h(x) = 2x + 1$
Solution $h(-1) = -1$

Problem	Solution
Ex 1. Is $f(x) = x^2$ continuous at $x = 2$ ?	Solution: $f(2) = 4$ , $\lim_{x \to 2} x^2 = 4$ ; they match, so continuous.
Ex 2. Is $g(x) = \frac{1}{x}$ continuous at $x = 0$ ?	Solution: $g(0)$ is not defined → discontinuous.
Ex 3. Is $h(x) = 2x + 1$ continuous at $x = -1$ ?	Solution: $h(-1) = -1$ , $\lim_{x \to -1} (2x+1) = -1$ ; they match, so continuous.

Problem types and how to solve

Type $Polynomial limit$
Description $Continuous so limit = function value$
How to get the answer $x=a$

Type $Linear limit$
Description $Same; limit equals function value$
How to get the answer $x=a$

Type $Continuous? (1/0)$
Description $1 if continuous at that point, 0 if not$
How to get the answer $f(a)$

Type $Limit at a hole$
Description $Limit at a point where there is a hole$
How to get the answer $x$

Type	Description	How to get the answer
Polynomial limit	Continuous so limit = function value	Substitute $x=a$ .
Linear limit	Same; limit equals function value	Substitute $x=a$ .
Continuous? (1/0)	1 if continuous at that point, 0 if not	Check ① $f(a)$ exists ② limit exists ③ limit $= f(a)$ .
Limit at a hole	Limit at a point where there is a hole	Use the formula (excluding the point) and let $x$ approach that point; if you can substitute, do so.

Example (polynomial · limit)

f(x)=x^2

continuous at

x=2

Solution

f (2) =4

\lim_{x \to 2} x^2=4

; they match, so continuous. → Continuous

(1)

Example (discontinuous)

g(x)=\frac{1}{x}

continuous at

x=0

Solution

g(0)

is not defined (division by zero). → Discontinuous (0)

Example (linear · continuous)

h(x)=2x+1

continuous at

x=-1

Solution

h(-1)=-1

\lim_{x \to -1}(2x+1)=-1

; they match, so continuous. → Continuous

(1)

What is continuity?

Continuity is often described intuitively as a graph you can draw in one stroke without lifting the pencil. Mathematically, we need a precise condition: the limit as

x

approaches

a

must exist and equal the function value

f(a)

a

. We write

\lim_{x \to a} f(x) = f(a)

Three-step checklist for continuity at a point

x=a

(1) Function value exists:

f(a)

must be defined (no hole).

(2) Limit exists:

\lim_{x \to a} f(x)

must exist. The left limit (as

x

approaches

a

from below) and right limit (from above) must be equal; if one is 0 and the other 1 (like a step), the limit does not exist and the function is discontinuous.

(3) They match: the limit and

f(a)

must be equal (the path exists but the bridge must not be in the wrong place).

To check continuity at a point, verify: does $\lim_{x \to a} f(x)$ exist?, is $f(a)$ defined?, and are they equal?

Checklist:

①

f(a)

exists

②

\lim_{x \to a} f(x)

exists

③ limit

= f(a)

. If any fails, the function is discontinuous at that point.

Example problems and solutions are in the table below.

Problem $f(x) = x^2$
Solution $f(2) = 4$

Problem $g(x) = \frac{1}{x}$
Solution $g(0)$

Problem $h(x) = 2x + 1$
Solution $h(-1) = -1$

Problem	Solution
Ex 1. Is $f(x) = x^2$ continuous at $x = 2$ ?	Solution: $f(2) = 4$ , $\lim_{x \to 2} x^2 = 4$ ; they match, so continuous.
Ex 2. Is $g(x) = \frac{1}{x}$ continuous at $x = 0$ ?	Solution: $g(0)$ is not defined → discontinuous.
Ex 3. Is $h(x) = 2x + 1$ continuous at $x = -1$ ?	Solution: $h(-1) = -1$ , $\lim_{x \to -1} (2x+1) = -1$ ; they match, so continuous.

Problem types and how to solve

Type $Polynomial limit$
Description $Continuous so limit = function value$
How to get the answer $x=a$

Type $Linear limit$
Description $Same; limit equals function value$
How to get the answer $x=a$

Type $Continuous? (1/0)$
Description $1 if continuous at that point, 0 if not$
How to get the answer $f(a)$

Type $Limit at a hole$
Description $Limit at a point where there is a hole$
How to get the answer $x$

Type	Description	How to get the answer
Polynomial limit	Continuous so limit = function value	Substitute $x=a$ .
Linear limit	Same; limit equals function value	Substitute $x=a$ .
Continuous? (1/0)	1 if continuous at that point, 0 if not	Check ① $f(a)$ exists ② limit exists ③ limit $= f(a)$ .
Limit at a hole	Limit at a point where there is a hole	Use the formula (excluding the point) and let $x$ approach that point; if you can substitute, do so.

Example (polynomial · limit)

f(x)=x^2

continuous at

x=2

Solution

f (2) =4

\lim_{x \to 2} x^2=4

; they match, so continuous. → Continuous

(1)

Example (discontinuous)

g(x)=\frac{1}{x}

continuous at

x=0

Solution

g(0)

is not defined (division by zero). → Discontinuous (0)

Example (linear · continuous)

h(x)=2x+1

continuous at

x=-1

Solution

h(-1)=-1

\lim_{x \to -1}(2x+1)=-1

; they match, so continuous. → Continuous

(1)