Limit laws — sum, product, quotient, constant multiple: Worked examples

#	Case class	What makes it different	Example
C1	Polynomial — plug in	Bottom is never a worry; sum + const-mult + power	Ex 1
C2	Quotient, $M \neq = 0$	Denominator heads to a nonzero number — law fires cleanly	Ex 2
C3	Quotient $\frac{0}{0}$ , factorable	Law fails; cancel a common factor	Ex 3
C4	Quotient $\frac{0}{0}$ , needs rationalising	Square roots hide the common factor	Ex 4
C5	$\frac{( nonzero )}{0}$ — one-sided split	Law fails, but the sign of the bottom decides $\pm \infty$	Ex 5
C6	Product where one piece has no limit	Sum/product law forbidden; need Squeeze theorem	Ex 6
C7	Sum where each piece diverges but the sum is fine	Cannot split — Mistake B live	Ex 7
C8	Real-world word problem	Translate units, then apply a law	Ex 8
C9	Exam twist — solve for an unknown constant	Reverse-engineer so the law can fire	Ex 9

C2 — Quotient with a safe (nonzero) denominator

C3 — Quotient that gives $\frac{0}{0}$ but factors

Worked example Example 3 (cell C3)

Find $x \to 3 lim \frac{x ^{2} - 9}{x - 3}$ .

Forecast: try to plug in $x = 3$ first. What breaks?

Test the bottom: $lim_{x \to 3} (x - 3) = 0$ , and the top $\to 9 - 9 = 0$ too. So this is the indeterminate form $\tfrac00$. Why this step? The quotient law is forbidden here ( $M = 0$ ). $\frac{0}{0}$ does not mean "no limit" — it means "the law can't see the answer; dig deeper."
Factor the top: $x^{2} - 9 = (x - 3) (x + 3)$ . Why this step? A $\frac{0}{0}$ almost always hides a shared factor. Exposing it lets us cancel the thing that was going to zero.
Cancel — legally: $\frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3, for x \neq = 3.$ Why this step? A limit only cares about $x$ near $3$ , never $x = 3$ itself, so $x - 3 \neq = 0$ and dividing it out is honest.
Now sum law: $lim_{x \to 3} (x + 3) = 6$ .

Verify: table nearby $x = 2.999 \to 5.999$ , $x = 3.001 \to 6.001$ . Both squeeze on $6$ . ✓ The graph is the line $y = x + 3$ with a single hole at $(3, 6)$ — see the figure.

Look at the red open circle: the function has no value there, but the limit is the height the line is aiming for — exactly $6$ .

C4 — $\frac{0}{0}$ that needs rationalising

Worked example Example 4 (cell C4)

Find $x \to 0 lim \frac{x + 4 - 2}{x}$ .

Forecast: plug in $x = 0$ : top is $4 - 2 = 0$ , bottom is $0$ . Same $\frac{0}{0}$ trap — but there's no obvious factor to cancel because of the square root.

Confirm the form: top $\to 0$ , bottom $\to 0$ , so quotient law is off-limits.
Multiply top and bottom by the conjugate $x + 4 + 2$ : $\frac{x + 4 - 2}{x} \cdot \frac{x + 4 + 2}{x + 4 + 2} .$ Why this step? $(u - 2) (u + 2) = u - 4$ kills the square root and manufactures the hidden common factor. This is "rationalising" — the root version of factoring.
The top becomes $(x + 4) - 4 = x$ : $\frac{x}{x ( x + 4 + 2 )} = \frac{1}{x + 4 + 2}, x \neq = 0.$ Why this step? The $x$ that was forcing $\frac{0}{0}$ cancels, leaving a piece whose bottom is now safely nonzero.
Apply the quotient law (bottom $\to 4 + 2 = 4 \neq = 0$ ): $\frac{1}{0 + 4 + 2} = \frac{1}{4} .$

Verify: $x = 0.0001$ : $\frac{4.0001 - 2}{0.0001} \approx 0.24999 \dots \approx \frac{1}{4}$ . ✓

C5 — $\frac{nonzero}{0}$ : the sign of the bottom rules

Worked example Example 5 (cell C5 — all sub-cases)

Investigate $x \to 2 lim \frac{1}{x - 2}$ .

Forecast: top heads to $1$ (nonzero), bottom heads to $0$ . This is not $\frac{0}{0}$ — it's $\frac{1}{0}$ . Guess: does a single limit exist?

Bottom $\to 0$ , so the quotient law fails. But top $\to 1 \neq = 0$ , so this cannot cancel to a nice number — it must blow up. The only question left: toward $+ \infty$ or $- \infty$ ? Why this step? When only the bottom hits zero, the magnitude explodes; the two sides can disagree in sign, so we must split.
Right side $x \to 2^{+}$ (so $x > 2$ ): $x - 2$ is a tiny positive number, e.g. $0.001$ . Then $\frac{1}{0.001} = + 1000$ . $lim_{x \to 2^{+}} \frac{1}{x - 2} = + \infty.$
Left side $x \to 2^{-}$ (so $x < 2$ ): $x - 2$ is a tiny negative number, e.g. $- 0.001$ . Then $\frac{1}{- 0.001} = - 1000$ . $lim_{x \to 2^{-}} \frac{1}{x - 2} = - \infty.$
The two one-sided limits disagree, so the two-sided limit does not exist.

Verify: $x = 2.001 \to + 1000$ , $x = 1.999 \to - 1000$ . ✓ Opposite signs, unbounded — see the vertical asymptote in the figure.

C6 — Product where one factor has no limit

Worked example Example 6 (cell C6)

Find $x \to 0 lim x sin \frac{1}{x}$ .

Forecast: the piece $sin \frac{1}{x}$ oscillates forever faster as $x \to 0$ — it has no limit. Does the product still settle?

You may not use the product law. It needs both limits to exist, and $lim_{x \to 0} sin \frac{1}{x}$ does not exist. Why this step? Mistake B in reverse — a missing piece limit voids the law.
Bound the wild factor: for every input, $- 1 \leq sin \frac{1}{x} \leq 1$ . Multiply through by $∣ x ∣$ : $- ∣ x ∣ \leq x sin \frac{1}{x} \leq ∣ x ∣.$ Why this step? We can't track the wobble, but we can trap it between two tame walls that both go to $0$ . This is the Squeeze theorem.
Both walls $lim_{x \to 0} (- ∣ x ∣) = 0$ and $lim_{x \to 0} ∣ x ∣ = 0$ . Squeezed: $lim_{x \to 0} x sin \frac{1}{x} = 0.$

Verify: $x = 0.001$ : value $= 0.001 \cdot sin (1000)$ , magnitude $\leq 0.001$ . ✓ Trapped inside a shrinking corridor — the figure shows the two straight walls $\pm ∣ x ∣$ pinching the oscillation to zero.

Forecast: the bottom heads to $0$ . For a finite limit, what must the top do?

Bottom $\to 0$ . A finite limit is only possible if the top also $\to 0$ — otherwise it's the $\frac{nonzero}{0}$ blow-up of cell C5. Why this step? We engineer a $\frac{0}{0}$ on purpose so a common factor can cancel and rescue the limit.
Force the top to vanish at $x = 5$ : substitute $x = 5$ into $x^{2} + k x - 5$ and set it to $0$ : $25 + 5 k - 5 = 0 \Rightarrow 5 k = - 20 \Rightarrow k = - 4.$ Why this step? If $(x - 5)$ is to be a factor of the top, the top must equal $0$ at $x = 5$ (factor theorem).
Check it cancels. With $k = - 4$ : top $= x^{2} - 4 x - 5 = (x - 5) (x + 1)$ . So $\frac{( x - 5 ) ( x + 1 )}{x - 5} = x + 1, x \neq = 5.$
Now the limit exists: $lim_{x \to 5} (x + 1) = 6$ .

Verify: $k = - 4$ , $x = 5.001$ : $\frac{( 5.001 ) ^{2} - 4 ( 5.001 ) - 5}{0.001} \approx 6.001$ . ✓ The limit equals $6$ , and it exists precisely because we tuned $k$ .

Connections

Limit laws — sum, product, quotient, constant multiple — the parent rules these examples exercise.
Indeterminate forms 0 over 0 — cells C3, C4, C9 all resolve this form.
One-sided limits — the engine of cell C5.
Squeeze theorem — the rescue in cell C6 when the product law is void.
Continuity — why C2 and C8 collapse to plug-in.
Epsilon-Delta definition of a limit — the ultimate justification behind every "cancel then substitute."
Derivative as a limit — cell C4's rationalising trick is exactly how you differentiate $x$ .

Worked examples — Limit laws — sum, product, quotient, constant multiple

The scenario matrix

C1 — Polynomial: just plug in

C2 — Quotient with a safe (nonzero) denominator

C3 — Quotient that gives $\frac{0}{0}$ but factors

C4 — $\frac{0}{0}$ that needs rationalising

C5 — $\frac{nonzero}{0}$ : the sign of the bottom rules

C6 — Product where one factor has no limit

C7 — Sum where each piece diverges but the total is fine

C8 — Real-world word problem

C9 — Exam twist: make the law able to fire

Connections