Functions, limits, and continuity: Question bank

Before we start, let's earn every symbol and picture this page will lean on.

Notation earned first

Figure — Functions, limits, and continuity

Now the three words this page stress-tests:

True or false — justify

A function can have two different inputs give the same output.

True. "Exactly one output per input" is a one-way rule;

f (x) = x^{2}

sends both

- 2

and

2

4

, and that is perfectly legal.

lim_{x \to c} f (x) = L

then

f (c) = L

False. The limit ignores the point

x = c

;

f (c)

can be missing or set to any wrong value (a hole), yet the limit is still

L

. Equality only holds when

f

is continuous at

c

f (x) = \frac{x ^{2} - 1}{x - 1}

and

g (x) = x + 1

are the same function.

False. They agree everywhere except

x = 1

, where

f

is undefined (

0/0

) but

g (1) = 2

. Different domains means different functions, even though their limits at

1

both equal

2

Every continuous function is differentiable.

False. Continuity forbids jumps and holes but not corners; ReLU

max (0, x)

and

∣ x ∣

are continuous everywhere yet have no single slope at

0

Every differentiable function is continuous.

True. Having a well-defined slope requires the graph to arrive exactly where it heads — a jump or hole would make the difference quotient blow up, so differentiability forces continuity (see Derivatives and gradients).

A limit that produces

0/0

does not exist.

False.

0/0

is an indeterminate form, a signal to simplify — factor, rationalize, or use L'Hôpital's Rule. The limit may exist perfectly well (e.g. it is

2

for

\frac{x ^{2} - 1}{x - 1}

1

If both one-sided limits exist, the two-sided limit exists.

False. They must also be equal. A step function has a left limit

lim_{x \to c^{-}} f

and a right limit

lim_{x \to c^{+}} f

at the step, but they differ, so the two-sided limit does not exist (a jump discontinuity).

Redefining a single point can turn a discontinuous function into a continuous one.

True — but only for removable discontinuities. If the limit exists, set

f (c)

equal to it and the hole is patched. Jumps and infinite blow-ups cannot be fixed by moving one point.

y^{2} = x

defines

y

as a function of

x

False. At

x = 4

it gives both

y = + 2

and

y = - 2

; a vertical line at

x = 4

cuts the curve twice, failing the "exactly one output" rule.

Continuity of

f

and

g

guarantees continuity of

f \circ g

(their composition).

True. A chain of continuous machines stays continuous — this is the backbone of backprop; see Continuity of composite functions.

Spot the error

"To find

lim_{x \to 1} \frac{x ^{2} - 1}{x - 1}

, plug in

x = 1

: you get

0/0

, so the limit is undefined."

The error is treating

0/0

as a verdict. It is a prompt to factor:

\frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1

for

x \neq = 1

, giving limit

2

f (x) = 1/ x

is continuous because its formula has no gaps."

The formula is fine, but

0

is not in the domain and

f \to \pm \infty

there — an infinite discontinuity. Continuity is only claimed on the domain; you cannot ignore where the machine has no output.

"The limit as

x \to 3

2 x + 1

7

because we plugged in

3

The value

7

is correct, but plugging in is only valid because

2 x + 1

is continuous — a conclusion, not the definition. The honest route: we want the output-gap

∣2 x + 1 - 7∣ = ∣2 x - 6∣ = 2∣ x - 3∣

below any

ε

; since it equals

2 \times

(input-gap), forcing the input-gap under

ε /2

makes the output-gap under

ε

— hence choose

δ = ε /2

"ReLU has a jump at

0

, so gradient descent breaks."

ReLU has no jump — its two one-sided limits at

0

both equal

0

(continuous). It has a corner (non-differentiability), handled by a subgradient, so Gradient Descent still works.

"Since

x = 4

maps to a single output under

y = x

, and so does

x = 9

, the square-root relation with

\pm

is also a function."

The

\pm x

relation gives two outputs per input, failing the vertical line test. Only the positive-root choice

y = x

is a function.

"A hole in the graph means the limit doesn't exist there."

A removable hole means

f (c)

is missing, but the two-sided limit can still exist and equal the height of the hole — that is exactly the removable case.

Why questions

Why does the limit definition use the strict

0 < ∣ x - c ∣

instead of

∣ x - c ∣

The strict

0 <

excludes

x = c

itself, encoding "we track the journey, not the arrival". Without it, a bad value at

c

could wreck a perfectly good limit.

Why does ML insist on continuous (and mostly differentiable) loss functions?

Gradient Descent follows the slope downhill; a jump has no meaningful slope and a hole has no value, so the ball can't roll smoothly — continuity is what makes the slope trustworthy.

Why is factoring allowed inside a

0/0

limit even though it cancels the term that is zero?

Because the limit never evaluates at

x = c

, only near it, so

x - c \neq = 0

throughout — dividing by it is legal on the domain the limit actually sees.

Why is "the vertical line test" the right test for functionhood?

A vertical line fixes one input

x

and reads off every output the curve gives there; two hits mean two outputs for one input, violating "exactly one output".

Why is continuity weaker than differentiability?

Continuity only asks the graph to arrive where it heads (no jump/hole); differentiability additionally demands a single slope, which corners like ReLU's kink at

0

fail to provide.

Why can a jump discontinuity never be fixed by redefining one point?

The left and right limits disagree, so no single value can equal both headings at once; you'd need to move a whole side, not one point.

Edge cases

Is a constant function

f (x) = 5

continuous everywhere?

Yes. For every

c

f (c) = 5

and

lim_{x \to c} f (x) = 5

; the heading and arrival always match, so the pen never lifts.

What is

lim_{x \to 0} \frac{∣ x ∣}{x}

Does not exist. From the right

lim_{x \to 0^{+}} = + 1

, from the left

lim_{x \to 0^{-}} = - 1

; the one-sided limits disagree, so the two-sided limit fails — a jump.

What is

lim_{x \to 0} sin (1/ x)

, and which break-type is it?

Does not exist — an oscillatory (essential) discontinuity. As

x \to 0

the input

1/ x

races through infinitely many cycles, so

sin (1/ x)

keeps sweeping between

- 1

and

+ 1

forever, never settling on a heading. It neither jumps nor blows up.

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

continuous at every point of its domain?

Yes. On its domain (all

x \neq = 0

) it has no breaks;

x = 0

simply isn't in the domain, so it isn't a discontinuity of the function — it's outside where the machine operates.

At the corner of ReLU (

x = 0

), what exists and what doesn't?

The value

f (0) = 0

and the limit

0

both exist and agree (continuous), but the slope does not exist uniquely (left slope

0

, right slope

1

) — non-differentiable.

Can a function be defined at

c

, have a limit at

c

, yet still be discontinuous at

c

Yes. If

f (c)

exists and

lim_{x \to c} f

exists but they are unequal (a misplaced point), condition 3 fails — a removable discontinuity.

What happens to continuity at an endpoint of a closed interval, say

x = 0

for

x

[0, \infty)

We use the one-sided limit:

lim_{x \to 0^{+}} x = 0 = 0

, so

x

is (right-)continuous at the endpoint even though no left side exists.

Recall One-line survival guide

Approach ≠ Arrive. Limit = where you're heading; value = where you land; continuous = they match. $0/0$ = simplify, not surrender. Continuous is weaker than differentiable. Two-sided limit exists only when both one-sided doors agree. Always check domain before declaring continuity.

Connections

Derivatives and gradients — why differentiable ⇒ continuous.
Gradient Descent — why smoothness matters for rolling downhill.
Activation functions (ReLU, sigmoid) — the continuous-but-not-differentiable star.
L'Hôpital's Rule — the tool for stubborn $0/0$ and $\infty/\infty$ .
Continuity of composite functions — chains of continuous machines stay continuous.