Limit laws — sum, product, quotient, constant multiple: Question bank

True or false — justify

Every convergent limit law needs the parts' limits to exist first.

True. The statement "if

lim f = L

and

lim g = M

then..." is an implication; when the hypotheses fail, the conclusion is simply not claimed — see Mistake B in the parent.

lim (f + g)

existing forces both

lim f

and

lim g

to exist.

False. Counterexample:

f = ⌊ x ⌋

g = - ⌊ x ⌋

at an integer — each jumps and has no limit, yet

f + g \equiv 0

has limit

0

lim (f g)

existing forces

lim f

and

lim g

each to exist.

False. Take

f (x) = x

and

g (x) = sin (1/ x)

near

0

g

has no limit, but

f g \to 0

by the Squeeze theorem. The product can converge while a factor does not.

lim f = L

exists but

lim g

does not, then

lim (f + g)

cannot exist.

True. If

lim (f + g)

existed, then

g = (f + g) - f

would have a limit by the difference law — contradiction. So the sum inherits

g

's non-existence.

The quotient law

lim (f / g) = L / M

holds whenever

M \neq = 0

, regardless of

L

True.

M \neq = 0

is the only condition;

L

may be anything, including

0

(then the quotient limit is just

0

M = 0

then

lim (f / g)

must fail to exist.

False. If

L = 0

too you get the indeterminate

0/0

, which can still have a finite limit after factoring — e.g.

\frac{x ^{2} - 9}{x - 3} \to 6

. See Indeterminate forms 0 over 0.

The power law

lim [f]^{n} = L^{n}

requires

n

to be a positive integer.

True for the proof by repeated product. For general real exponents you need continuity of

x \mapsto x^{n}

and a positivity/limit condition, not just the product law.

Direct substitution

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

works for every function, not just polynomials.

False. It works exactly when the function is continuous at

a

. Polynomials happen to be continuous everywhere; a function with a hole or jump at

a

breaks it.

The limit laws apply just as well to One-sided limits.

True. Every proof only used "

x

near

a

"; restricting to

x < a

x > a

changes nothing in the triangle-inequality argument, so all four laws hold one-sidedly.

lim (c f) = c L

still holds when

c = 0

True. Both sides equal

0

0 \cdot f = 0

and

c L = 0 \cdot L = 0

, no matter what

L

is (even if

L

existed as a finite value).

Spot the error

lim_{x \to 3} \frac{x - 3}{x - 3} = \frac{0}{0}

, which is undefined, so the limit doesn't exist."

Wrong conclusion. The limit only sees

x

near 3 where

x \neq = 3

, so the ratio is

1

throughout and the limit is

1

; the value at

x = 3

is irrelevant to a limit.

lim (f \cdot g) = lim f \cdot lim g

, so

lim (x \cdot \frac{1}{x}) = lim x \cdot lim \frac{1}{x} = 0 \cdot \infty

Illegal use.

lim_{x \to 0} \frac{1}{x}

does not exist (finite), so the product law's hypothesis fails — you cannot apply it. Simplify first:

x \cdot \frac{1}{x} = 1 \to 1

"By the quotient law

lim_{x \to 0} \frac{s i n x}{x} = \frac{l i m s i n x}{l i m x} = \frac{0}{0} = 1

Two errors:

0/0

is not a number, and the quotient law is void when

M = 0

. This limit equals 1 but needs the Squeeze theorem or Derivative as a limit, not the quotient law.

lim (f + g) = L + M

, so I can add limits even if only one of them exists."

The law names both

L

and

M

as existing finite limits. With one missing there is no

M

to add, so the equation is meaningless as stated.

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

works, so

lim_{x \to a} (g \cdot f) = g \cdot L

for any

g

The constant-multiple law needs a genuine constant

c

, not a varying

g (x)

. Replacing the constant by a function requires the full product law and

lim g

to exist.

"In the sum proof we chose

∣ f - L ∣ < ε

and

∣ g - M ∣ < ε

, giving total

< ε

Adding two

ε

's gives

2 ε

, not

ε

. That's why the parent splits into

ε /2 + ε /2

— so the halves reassemble to exactly

ε

Why questions

Why does the product-law proof add and subtract

L g (x)

It splits the total wobble into

g (f - L) + L (g - M)

— one piece isolates

f

's error, the other

g

's error — and each piece is a known controllable limit heading to

0

Why must

g

be bounded near

a

for the product proof to close?

g (f - L)

, if

g

blew up it could amplify a tiny

∣ f - L ∣

into something large. Because

g \to M

, it stays within a fixed band near

a

, so

g (f - L) \to 0

Why is

M \neq = 0

the fine print on the quotient law and nowhere else?

Dividing by a quantity heading to

0

can send

f / g

\pm \infty

or make it oscillate; only away-from-zero denominators stay finite, letting

1/ g \to 1/ M

be proved.

Why can we compute polynomial limits by plain substitution?

A polynomial is sums of constant-multiples of powers of

x

; each atom obeys

lim x = a

and

lim c = c

, and the sum/product/const-multiple laws reassemble them into

p (a)

Why do we "almost never" use epsilon–delta directly to find a limit?

The laws let us decompose a messy function into atoms with known limits and glue the answers together, so one general proof of each law replaces countless case-by-case

ε

δ

hunts.

Why does the triangle inequality appear in the sum proof and not, say, the product proof's first line?

The sum's error is literally

(f - L) + (g - M)

, a sum of two terms — the exact shape

∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣

bounds. The product's error is a sum only after the add-subtract trick creates it.

Edge cases

If both

L = 0

and

M = 0

, what does the quotient law tell you?

Nothing — it doesn't apply (

M = 0

). You face

0/0

, an Indeterminate forms 0 over 0; the true limit may be any number,

\pm \infty

, or nonexistent.

L \neq = 0

but

M = 0

, can

lim (f / g)

be finite?

Generally no — numerator heads to a nonzero value while the denominator vanishes, so

∣ f / g ∣ \to \infty

(or the two-sided limit fails from a sign flip). It is not an indeterminate form.

At a point where

lim_{x \to a^{-}} f \neq = lim_{x \to a^{+}} f

, do the sum/product laws still work?

Yes, but only for the one-sided limits that exist. The two-sided

lim_{x \to a} f

doesn't exist, so you may only combine matching one-sided limits via One-sided limits.

Can the product of two functions with no limit have a limit?

Yes.

f = sgn (x)

and

g = sgn (x)

near

0

each jump, but

f g \equiv 1

(for

x \neq = 0

) has limit

1

. Non-existence of factors does not forbid a product limit.

Does

lim (c f) = c L

still make sense if

L

+ \infty

The stated law assumes

L

finite. With

L = + \infty

you leave the finite-limit laws entirely; extended-real rules apply (and

c = 0

then gives the indeterminate

0 \cdot \infty

What happens to the sum law if one limit is

+ \infty

and the other is

- \infty

The finite law does not cover it;

\infty - \infty

is indeterminate. You must analyse the specific functions directly — the algebraic law gives no answer.

Connections

Limit laws — sum, product, quotient, constant multiple — the parent whose fine print these traps probe.
Epsilon-Delta definition of a limit — the source of every "why" answer above.
One-sided limits — where mismatched left/right limits break two-sided combination.
Indeterminate forms 0 over 0 — the destination when the quotient law is voided.
Continuity — why direct substitution works only for continuous functions.
Squeeze theorem — rescues product/quotient limits the algebra laws can't reach.
Derivative as a limit — where these same traps reappear in the difference quotient.