One-sided limits — left-hand, right-hand: Worked examples

The scenario matrix

Every one-sided-limit question is really one of a handful of shapes. Here is the full list, and which example kills each.

Cell	Case class	What makes it tricky	Example
A	Sides agree (piecewise)	must check both rules	Ex 1
B	Sides disagree, both finite (jump)	equal-existence trap	Ex 2
C	Domain edge — one side undefined	only ONE side exists	Ex 3
D	Pole / infinite — sides go to $\pm \infty$	sign of denominator	Ex 4
E	Floor / step, tricky non-integer twist	floor jumps only at integers	Ex 5
F	$0/0$ removable — algebra needed first	can't just plug in	Ex 6
G	Squeeze at a wiggle ( $sin \frac{1}{x}$ type)	bounded-times-small	Ex 7
H	Word problem (price/tax jump)	translate story → piecewise	Ex 8
I	Exam twist — solve for an unknown constant	force sides to match	Ex 9

We build the vocabulary from zero first, then march through A→I.

Ground rules, restated in plain words

Figure — One-sided limits — left-hand, right-hand

The figure above is our map: a single number line with $a$ marked. The coral arrow creeps in from the right ( $a^{+}$ ), the lavender arrow from the left ( $a^{-}$ ). Every example below is just "what value is each arrow's function heading toward?"

Cell A — sides agree (piecewise)

Cell B — sides disagree, both finite (a jump)

Worked example Ex 2 · The classic sign-jump

$s (x) = \frac{∣ x - 4∣}{x - 4}, x \neq = 4. Find both one-sided limits at 4.$

Forecast: absolute value over the thing inside — this is a " $+ 1$ / $- 1$ " splitter. Guess: right $= + 1$ , left $= - 1$ , no two-sided limit.

Right side ( $x > 4$ ): then $x - 4 > 0$ , so $∣ x - 4∣ = x - 4$ . Ratio $= \frac{x - 4}{x - 4} = 1$ . Why this step? The definition of absolute value: for a positive quantity, $∣ u ∣ = u$ . Here $u = x - 4$ is positive when $x > 4$ .
So $lim_{x \to 4^{+}} s = 1$ .
Left side ( $x < 4$ ): then $x - 4 < 0$ , so $∣ x - 4∣ = - (x - 4)$ . Ratio $= \frac{- ( x - 4 )}{x - 4} = - 1$ . Why this step? For a negative quantity, $∣ u ∣ = - u$ (flip the sign to make it positive).
So $lim_{x \to 4^{-}} s = - 1$ . Since $1 \neq = - 1$ , the two-sided limit does not exist — a jump of size $2$ .

Verify: $s (4.5) = \frac{∣0.5∣}{0.5} = 1$ ✓ and $s (3.5) = \frac{∣ - 0.5∣}{- 0.5} = \frac{0.5}{- 0.5} = - 1$ ✓.

Cell C — domain edge (only one side lives)

Worked example Ex 3 · A square-root wall

Find the one-sided limits of $r (x) = x - 2$ at $x = 2$ .

Forecast: you can't square-root a negative real. So the left side has nowhere to stand. Guess: only the right limit exists, and it's $0$ .

Domain check first. $x - 2$ is real only when $x - 2 \geq 0$ , i.e. $x \geq 2$ . Why this step? A limit can only "approach" through points where the function is defined. No domain on the left = no left-hand limit to speak of.
Right side ( $x > 2$ ): as $x \to 2^{+}$ , $x - 2 \to 0^{+}$ , so $x - 2 \to 0$ . Thus $lim_{x \to 2^{+}} r = 0$ .
Left side ( $x < 2$ ): the function is undefined; $lim_{x \to 2^{-}} r$ does not exist (no domain).
Two-sided: does not exist, because one side is missing — but for a purpose-built reason: it's a domain edge, not a mismatch.

Verify: $r (2.01) = 0.01 = 0.1$ , and $r (2.0001) = 0.0001 = 0.01$ — shrinking toward $0$ ✓. $r (1.9) = - 0.1$ is not a real number ✓.

Cell D — pole / infinite behaviour

Cell E — floor function, non-integer twist

Worked example Ex 5 · Floor where the approach point is NOT an integer

Find both one-sided limits of $⌊ 2 x ⌋$ at $x = 1.5$ .

Forecast: floor only jumps where its input hits an integer. Input is $2 x$ ; at $x = 1.5$ that's $2 (1.5) = 3$ , an integer — so a jump does happen here. Guess: left $= 2$ , right $= 3$ .

Find where the inside is integer. $2 x = 3$ exactly at $x = 1.5$ . So the "step" of the floor lands precisely on our point. Why this step? $⌊ \cdot ⌋$ is constant between integers and jumps at them; we must check whether $2 x$ crosses an integer at $x = 1.5$ . It does.
Left side ( $x < 1.5$ ): say $x = 1.49$ , then $2 x = 2.98$ , and $⌊ 2.98 ⌋ = 2$ . So $lim_{x \to 1. 5^{-}} ⌊ 2 x ⌋ = 2$ . Why this step? Just below $1.5$ , $2 x$ is just below $3$ , so it floors down to $2$ .
Right side ( $x > 1.5$ ): say $x = 1.51$ , then $2 x = 3.02$ , and $⌊ 3.02 ⌋ = 3$ . So $lim_{x \to 1. 5^{+}} ⌊ 2 x ⌋ = 3$ .
Two-sided: $2 \neq = 3$ , so it does not exist — a jump of $1$ .

Verify: $⌊ 2 (1.499)⌋ = ⌊ 2.998 ⌋ = 2$ ✓ and $⌊ 2 (1.501)⌋ = ⌊ 3.002 ⌋ = 3$ ✓.

Cell F — removable $0/0$ (algebra first)

Worked example Ex 6 · You cannot just plug in

Find $lim_{x \to 3^{-}}$ and $lim_{x \to 3^{+}}$ of $q (x) = \frac{x ^{2} - 9}{x - 3}$ .

Forecast: at $x = 3$ we'd get $\frac{0}{0}$ — undefined, so plugging in is illegal. But it factors. Guess: both sides give $6$ ; limit exists $= 6$ .

Diagnose. Direct substitution gives $\frac{3 ^{2} - 9}{3 - 3} = \frac{0}{0}$ , an indeterminate form — no information yet. Why this step? $\frac{0}{0}$ means "cannot read off"; we must simplify before taking the limit.
Factor the top. $x^{2} - 9 = (x - 3) (x + 3)$ (difference of squares). So $q (x) = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ for $x \neq = 3$ . Why this step? Cancelling the shared $(x - 3)$ removes the trouble; valid because $x \neq = 3$ near the limit (we never touch $x = 3$ ).
Now both sides use $x + 3$ . Left: $x \to 3^{-}$ , $x + 3 \to 6$ . Right: $x \to 3^{+}$ , $x + 3 \to 6$ .
Both give $6$ , equal, so $lim_{x \to 3} q = 6$ . (The hole at $x = 3$ is removable.)

Verify: $q (2.99) = \frac{2.9 9 ^{2} - 9}{2.99 - 3} = \frac{- 0.0599}{- 0.01} = 5.99$ ✓ and $q (3.01) = \frac{3.0 1 ^{2} - 9}{0.01} = 6.01$ ✓.

Cell G — squeeze at a wiggle

Worked example Ex 7 · Infinite oscillation, tamed

Find $lim_{x \to 0^{+}}$ and $lim_{x \to 0^{-}}$ of $w (x) = x sin (\frac{1}{x})$ .

Forecast: $sin (1/ x)$ oscillates crazily near $0$ , but $x$ shrinks to $0$ . A small number times a bounded number → $0$ . Guess: both sides $= 0$ .

Bound the wiggle. For every $x \neq = 0$ , $- 1 \leq sin (1/ x) \leq 1$ . Why this step? $sin$ of anything stays inside $[- 1, 1]$ ; we don't need to know where it is, only that it's bounded.
Multiply by $∣ x ∣$ . Then $- ∣ x ∣ \leq x sin (1/ x) \leq ∣ x ∣$ . Why this step? Multiplying the bound by $∣ x ∣$ traps $w (x)$ between two things that both go to $0$ — the squeeze.
Both bounds → 0. As $x \to 0$ from either side, $∣ x ∣ \to 0$ and $- ∣ x ∣ \to 0$ , so $w (x)$ is squeezed to $0$ .
Therefore $lim_{x \to 0^{-}} w = lim_{x \to 0^{+}} w = 0$ , and the two-sided limit is $0$ — even though $w$ oscillates infinitely often.

Verify: $w (0.01) = 0.01 sin (100) \approx 0.01 (- 0.506) = - 0.00506$ , tiny ✓. $w (- 0.01) = - 0.01 sin (- 100) \approx - 0.00506$ , also tiny ✓. Both near $0$ .

Cell H — word problem (translate the story)

Worked example Ex 8 · A parking-garage price jump

A garage charges $3 per hour, but the moment you pass $2$ hours you are billed a flat $8 for the whole session up to $4$ hours. Model the cost $C (t)$ and find the one-sided limits at $t = 2$ hours.

Forecast: below $2$ hours cost climbs toward $3\times2=\$ 6 $; a t / a f t er$ 2 $hours it jumps to \$ 8. Guess: left $= 6$ , right $= 8$ , no two-sided limit — the customer sees a price cliff.

Translate to a piecewise rule. $C (t) = {3 t 8 0 \leq t < 2 2 \leq t \leq 4$ Why this step? "Per-hour up to 2h" is the line $3 t$ ; "flat 8 after passing 2h" is a constant.
Left side ( $t \to 2^{-}$ ): use $3 t$ : $3 (2) = 6$ . So $\lim_{t\to2^-}C=\$ 6 $. * W h y t hi ss t e p ? * J u s t u n d er$ 2$ hours you're still on the per-hour meter.
Right side ( $t \to 2^{+}$ ): use $8$ : constant, so $\lim_{t\to2^+}C=\$ 8$.
Interpret: $6 \neq = 8$ , no two-sided limit — a real jump discontinuity of $2. That extra $2 is the "penalty" for crossing the 2-hour line.

Verify (units $): $C (1.99) = 3 (1.99) = 5.97 \approx 6$ ✓; $C (2.01) = 8$ ✓. Prices behave as forecast.

Cell I — exam twist: solve for the unknown

Worked example Ex 9 · Make the pieces meet

$f (x) = {k x + 2 x^{2} + k x < 1 x \geq 1$ For what value of $k$ does $lim_{x \to 1} f (x)$ exist? (This ties into Continuity at a point.)

Forecast: the two-sided limit exists only if the sides match; set them equal and solve for $k$ . Guess: one clean value of $k$ .

Left side ( $x \to 1^{-}$ ): rule $k x + 2$ gives $k (1) + 2 = k + 2$ . Why this step? Below $1$ the active rule is the line $k x + 2$ ; substitute the approached value $1$ .
Right side ( $x \to 1^{+}$ ): rule $x^{2} + k$ gives $1 + k$ .
Force equality (the law). For the two-sided limit to exist: $k + 2 = 1 + k ⟹ 2 = 1.$ Why this step? We demanded the sides match; but $2 = 1$ is false for every $k$ .
Conclusion: there is no value of $k$ that makes the limit exist — the constant terms clash no matter what. (A sneaky exam trap: the unknown you're solving for cancels out.)

Verify: subtract the two expressions: $(k + 2) - (1 + k) = 1 \neq = 0$ , independent of $k$ ✓. The sides always differ by exactly $1$ .

Wrap-up: the matrix, filled

Recall Which example handled which shape?

A agree ::: Ex 1 (limit $= 3$ ) B jump ::: Ex 2 ( $+ 1$ vs $- 1$ ) C domain edge ::: Ex 3 (right $= 0$ , left undefined) D pole ::: Ex 4 ( $+ \infty$ vs $- \infty$ ) E floor twist ::: Ex 5 ( $2$ vs $3$ ) F removable $0/0$ ::: Ex 6 (both $= 6$ ) G squeeze wiggle ::: Ex 7 (both $= 0$ ) H word problem ::: Ex 8 ($6 vs $8) I unknown constant ::: Ex 9 (no $k$ works)

Connections

One-sided Limits — Left-hand & Right-hand (parent)
Limit of a function — intuitive & ε-δ definition
Continuity at a point
Jump, removable & infinite discontinuities
Vertical asymptotes
Greatest integer / floor function
Differentiability — left & right derivatives

Worked examples — One-sided limits — left-hand, right-hand

The scenario matrix

Ground rules, restated in plain words

Cell A — sides agree (piecewise)

Cell B — sides disagree, both finite (a jump)

Cell C — domain edge (only one side lives)

Cell D — pole / infinite behaviour

Cell E — floor function, non-integer twist

Cell F — removable $0/0$ (algebra first)

Cell G — squeeze at a wiggle

Cell H — word problem (translate the story)

Cell I — exam twist: solve for the unknown

Wrap-up: the matrix, filled

Connections

Concept Map