Infinite limits and limits at infinity — vertical - horizontal asymptotes: Worked examples

#	Case class	What makes it distinct	Answer type	Example
A	Nonzero ÷ zero, sides disagree	left limit and right limit have opposite signs	VA, limit DNE	Ex 1
B	Nonzero ÷ zero, sides agree (squared denom)	denominator $\to 0^{+}$ both sides	VA, $\pm \infty$ both sides	Ex 2
C	Zero ÷ zero — removable	top and bottom share a factor	hole, NOT a VA	Ex 3
D	Rational at infinity, deg top $<$ deg bottom	numerator dies faster	HA $y = 0$	Ex 4
E	Rational at infinity, deg top $=$ deg bottom	leading terms survive	HA = ratio of leaders	Ex 4
F	Rational at infinity, deg top $>$ deg bottom	numerator wins	no HA ( $\pm \infty$ )	Ex 4
G	Root + sign care, $x \to - \infty$	$\sqrt{x^2}=	x	=-x$
H	Transcendental ( $e^{- x}$ , $arctan$ )	not rational, still settles	HA from known limits	Ex 6
I	Word problem (real-world settling)	asymptote = long-run value	HA with units	Ex 7
J	Exam twist — VA cancels but another survives	mixed hole + real asymptote	one hole, one VA	Ex 8
K	Slant asymptote (deg top $=$ deg bottom $+ 1$ )	numerator exactly one degree higher	oblique line $y = m x + b$	Ex 9

$x \to 1 lim \frac{x + 2}{x - 1}$ , and both one-sided limits. Forecast: Numerator at $x = 1$ is $3$ (nonzero), denominator $\to 0$ . So it blows up — but does it go up or down? Guess the sign on each side before reading on.

Step 1 — Check the numerator's sign near $x = 1$ . Why this step? If the top were also heading to $0$ we'd have a $\frac{0}{0}$ hole (cell C) and could NOT conclude an asymptote. Here $x + 2 \to 3 > 0$ , safely nonzero and positive. Good — this is a genuine "nonzero ÷ zero" (cell A/B).

Step 2 — Right side $x \to 1^{+}$ . Why? The sign of the answer is decided entirely by the denominator's sign, and that sign differs left vs right. Take $x = 1.001$ : $x - 1 = 0.001 > 0$ , a tiny positive. So $\frac{+ 3}{0 ^{+}} = + \infty$ .

Step 3 — Left side $x \to 1^{-}$ . Take $x = 0.999$ : $x - 1 = - 0.001 < 0$ , a tiny negative. So $\frac{+ 3}{0 ^{-}} = - \infty$ .

Step 4 — Combine. The two sides disagree ( $+ \infty$ vs $- \infty$ ), so $lim_{x \to 1} \frac{x + 2}{x - 1}$ does not exist. But $x = 1$ is still a vertical asymptote (a one-sided infinite limit is enough).

Verify: Plug $x = 1.0001$ : $\frac{3.0001}{0.0001} = 30001$ — huge positive ✓. Plug $x = 0.9999$ : $\frac{2.9999}{- 0.0001} = - 29999$ — huge negative ✓. Signs match Steps 2–3.

The figure below plots this function: watch the cyan curve rocket up just right of the amber line $x = 1$ and plunge down just left of it — that visible disagreement is exactly the " $+ \infty$ vs $- \infty$ " of Steps 2–3. The dotted white line is the horizontal asymptote $y = 1$ (equal degrees, ratio $1/1$ ).

Figure — Infinite limits and limits at infinity — vertical - horizontal asymptotes

Ex 2 — Cell B: squared denominator, sides agree

Worked example

$x \to - 2 lim \frac{5 - x}{( x + 2 ) ^{2}}$ Forecast: Denominator is a square, so it can't go negative. What does that force about the two sides? Guess before reading.

Step 1 — Numerator sign near $x = - 2$ . $5 - x \to 5 - (- 2) = 7 > 0$ , nonzero positive.

Step 2 — Denominator behaviour. Why a square matters: $(x + 2)^{2} \geq 0$ always, and it hits $0$ only at $x = - 2$ . Near $- 2$ it is a tiny positive — from both sides, because squaring kills the sign. So $(x + 2)^{2} \to 0^{+}$ left and right alike.

Step 3 — Combine. $\frac{+ 7}{0 ^{+}} = + \infty$ from both sides. Because the sides agree, the two-sided limit exists as $+ \infty$ : $lim_{x \to - 2} \frac{5 - x}{( x + 2 ) ^{2}} = + \infty.$ Vertical asymptote $x = - 2$ .

Verify: $x = - 1.99$ : $(0.01)^{2} = 0.0001$ , $\frac{6.99}{0.0001} = 69900$ ✓. $x = - 2.01$ : $(- 0.01)^{2} = 0.0001$ , $\frac{7.01}{0.0001} = 70100$ ✓. Both large positive — agreement confirmed.

Ex 3 — Cell C: the $\frac{0}{0}$ removable hole

Worked example

$x \to 4 lim \frac{x ^{2} - 16}{x - 4}$ — asymptote or hole? Forecast: Denominator $\to 0$ . Does it explode? Check the numerator first before guessing.

Step 1 — Test the numerator at $x = 4$ . $4^{2} - 16 = 0$ . Why this changes everything: both top and bottom $\to 0$ , so this is the indeterminate form $\frac{0}{0}$ — NOT automatically an asymptote. The vanishing factor might cancel.

Step 2 — Factor and cancel. Why factor? To expose and remove the shared $(x - 4)$ . $\frac{x ^{2} - 16}{x - 4} = \frac{( x - 4 ) ( x + 4 )}{x - 4} = x + 4 (x \neq = 4) .$ The cancellation is legal because in a limit $x \neq = 4$ , so we never divide by an actual zero.

Step 3 — Take the limit of the simplified form. $lim_{x \to 4} (x + 4) = 8$ . Finite! So there is no vertical asymptote — just a removable hole at the point $(4, 8)$ . (See Continuity and removable discontinuities.)

Verify: $x = 4.001 \Rightarrow x + 4 = 8.001 \approx 8$ ✓. Original at $x = 3.999$ : here $3.99 9^{2} = 15.992001$ , so $\frac{15.992001 - 16}{- 0.001} = \frac{- 0.007999}{- 0.001} = 7.999$ ✓ — no blow-up, confirming a hole not a wall.

Ex 4 — Cells D, E, F: rational functions at infinity (all three degree cases)

Worked example Evaluate all three at

$x \to \infty$ : $(D) \frac{2 x + 1}{x ^{2} + 3}, (E) \frac{4 x ^{2} - x}{2 x ^{2} + 9}, (F) \frac{x ^{3} + 1}{x ^{2}} .$ Forecast: One dies to $0$ , one lands on a nonzero number, one runs off to $\infty$ . Match each to its fate before reading.

The master move — divide by the highest denominator power. Why: this turns every term into a constant times $\frac{1}{x ^{k}}$ , and we know $lim_{x \to \infty} \frac{1}{x ^{k}} = 0$ for $k > 0$ . That single fact does all the work.

(D) deg top $1 <$ deg bottom $2$ . Divide by $x^{2}$ : $\frac{2 x + 1}{x ^{2} + 3} = \frac{\frac{2}{x} + \frac{1}{x ^{2}}}{1 + \frac{3}{x ^{2}}} \to \frac{0 + 0}{1 + 0} = 0.$ HA $y = 0$ . Why: the numerator only has negative powers of $x$ — it dies.

(E) deg top $=$ deg bottom $= 2$ . Divide by $x^{2}$ : $\frac{4 x ^{2} - x}{2 x ^{2} + 9} = \frac{4 - \frac{1}{x}}{2 + \frac{9}{x ^{2}}} \to \frac{4 - 0}{2 + 0} = 2.$ HA $y = 2$ — the ratio of leading coefficients $4/2$ .

(F) deg top $3 >$ deg bottom $2$ . Divide by $x^{2}$ : $\frac{x ^{3} + 1}{x ^{2}} = x + \frac{1}{x ^{2}} \to + \infty.$ No horizontal asymptote. (Since deg top exceeds deg bottom by exactly $1$ this is also a slant case — worked fully in Ex 9; general theory in Slant (oblique) asymptotes via polynomial division.)

Verify (plug $x = 1000$ ): (D) $\frac{2001}{1000003} \approx 0.0020 \to 0$ ✓. (E) $\frac{3999000}{2000009} \approx 1.9995 \to 2$ ✓. (F) $\frac{1 0 ^{9} + 1}{1 0 ^{6}} = 1000000.000001 \to \infty$ ✓.

Ex 5 — Cell G: root with sign care, two different asymptotes

Worked example Find

both horizontal asymptotes of $f (x) = \frac{9 x ^{2} + 4}{2 x - 1}$ . Forecast: A square root of $x^{2}$ hides a $∣ x ∣$ . Do you expect the same limit on both ends, or two different ones? Guess.

Step 1 — Pull $x^{2}$ out of the root. Why: to isolate a $\frac{1}{x ^{2}}$ term we can send to $0$ . $9 x^{2} + 4 = x^{2} 9 + \frac{4}{x ^{2}} = ∣ x ∣ 9 + \frac{4}{x ^{2}} .$ The key fact: $x^{2} = ∣ x ∣$ , not $x$ .

Step 2 — $x \to + \infty$ . Here $x > 0$ , so $∣ x ∣ = x$ . Divide top and bottom by $x$ : $\frac{x 9 + 4/ x ^{2}}{2 x - 1} = \frac{9 + 4/ x ^{2}}{2 - 1/ x} \to \frac{9}{2} = \frac{3}{2} .$ HA $y = \frac{3}{2}$ on the right.

Step 3 — $x \to - \infty$ . Why the sign flips: now $x < 0$ , so $∣ x ∣ = - x$ . That minus sign is the whole exam trap. $\frac{- x 9 + 4/ x ^{2}}{2 x - 1} = \frac{- 9 + 4/ x ^{2}}{2 - 1/ x} \to \frac{- 3}{2} = - \frac{3}{2} .$ HA $y = - \frac{3}{2}$ on the left.

Step 4 — Conclusion. Two different horizontal asymptotes: $y = \frac{3}{2}$ (right) and $y = - \frac{3}{2}$ (left).

Verify: $x = 1000$ : $\frac{9 , 000 , 004}{1999} \approx \frac{3000.0007}{1999} \approx 1.5008 \to 1.5$ ✓. $x = - 1000$ : $\frac{3000.0007}{- 2001} \approx - 1.4993 \to - 1.5$ ✓.

The figure makes the sign flip visible: the cyan curve levels off toward the amber line $y = \frac{3}{2}$ as you travel right, but toward the white line $y = - \frac{3}{2}$ as you travel left. That mirror-image split is entirely due to $∣ x ∣ = - x$ appearing on the left.

Ex 6 — Cell H: transcendental functions

Worked example Horizontal asymptotes of

$g (x) = \frac{e ^{x} - 1}{e ^{x} + 1}$ , both ends. Forecast: As $x$ grows, $e^{x}$ dwarfs the constants; as $x \to - \infty$ , $e^{x} \to 0$ . Two different ceilings? Guess.

Step 1 — $x \to + \infty$ . Why divide by $e^{x}$ : it's the fastest-growing term, the analogue of "highest power." $\frac{e ^{x} - 1}{e ^{x} + 1} = \frac{1 - e ^{- x}}{1 + e ^{- x}} .$ Since $e^{- x} = 1/ e^{x} \to 0$ : $\to \frac{1 - 0}{1 + 0} = 1.$ HA $y = 1$ on the right.

Step 2 — $x \to - \infty$ . Why NOT divide by $e^{x}$ here: as $x \to - \infty$ , $e^{x} \to 0$ , so plug it in directly. $\frac{e ^{x} - 1}{e ^{x} + 1} \to \frac{0 - 1}{0 + 1} = - 1.$ HA $y = - 1$ on the left.

Step 3 — Sanity on $arctan$ (parent's forecast drill). For comparison: $lim_{x \to \infty} arctan x = \frac{π}{2}$ and $lim_{x \to - \infty} arctan x = - \frac{π}{2}$ — same two-ceiling pattern.

Verify: $g (5) = \frac{e ^{5} - 1}{e ^{5} + 1} = \frac{147.41}{149.41} \approx 0.9866 \to 1$ ✓. $g (- 5) = \frac{e ^{- 5} - 1}{e ^{- 5} + 1} = \frac{- 0.9933}{1.0067} \approx - 0.9866 \to - 1$ ✓.

Ex 7 — Cell I: word problem (long-run value)

Worked example A drug's concentration in the blood (mg/L) after

$t$ hours is $C (t) = \frac{40 t}{t + 8}$ for $t \geq 0$ . What concentration does the patient approach after a very long time, and what does that number mean? Forecast: As hours pile up, does $C$ keep climbing forever or level off? Guess the ceiling.

Step 1 — Recognise the case. $C$ is a rational function, deg top $=$ deg bottom $= 1$ → this is cell E. Expect a finite horizontal asymptote = ratio of leading coefficients.

Step 2 — Take the limit at infinity. Divide by $t$ : $C (t) = \frac{40 t}{t + 8} = \frac{40}{1 + \frac{8}{t}} \to \frac{40}{1 + 0} = 40.$

Step 3 — Interpret with units. The horizontal asymptote is $C = 40$ mg/L. Physically: the concentration rises toward a plateau of 40 mg/L and never exceeds it — the long-run steady level. It never quite reaches $40$ (asymptotes are approached, not touched), matching how the graph hugs the ceiling.

Verify: $C (100) = \frac{4000}{108} \approx 37.04$ ; $C (10000) = \frac{400000}{10008} \approx 39.97 \to 40$ ✓. Rising toward $40$ , staying below it ✓. Units: $\frac{mg/L \cdot h}{h} =$ mg/L ✓.

The figure shows $C (t)$ climbing steeply at first, then bending to run flat just under the amber ceiling $C = 40$ — the curve approaches it but the amber arrow marks that it is never reached.

Ex 8 — Cell J: exam twist (a hole AND a real asymptote)

Worked example For

$h (x) = \frac{x ^{2} - 2 x - 3}{x ^{2} - 9}$ , find every vertical asymptote and every hole. Forecast: The denominator $x^{2} - 9 = (x - 3) (x + 3)$ vanishes at $x = 3$ and $x = - 3$ . Are both vertical asymptotes? Careful — check the numerator at each.

Step 1 — Factor both. Why: to see which zeros of the bottom are shared by the top (holes) and which are not (true asymptotes). $h (x) = \frac{( x - 3 ) ( x + 1 )}{( x - 3 ) ( x + 3 )} .$

Step 2 — At $x = 3$ : shared factor $(x - 3)$ cancels. Why this means a hole: both top and bottom hit $0$ there — a $\frac{0}{0}$ form that cancels, so no blow-up. Simplified: $h (x) = \frac{x + 1}{x + 3}$ for $x \neq = 3$ . The hole's height: plug $x = 3$ into the simplified form, $\frac{3 + 1}{3 + 3} = \frac{4}{6} = \frac{2}{3}$ . Hole at $(3, \frac{2}{3})$ .

Step 3 — At $x = - 3$ : factor does NOT cancel. In $\frac{x + 1}{x + 3}$ the denominator still vanishes at $x = - 3$ while the numerator gives $- 3 + 1 = - 2 \neq = 0$ . Nonzero ÷ zero → vertical asymptote $x = - 3$ . (Odd power → sides will disagree: left is $\frac{- 2}{0 ^{-}} = + \infty$ , right is $\frac{- 2}{0 ^{+}} = - \infty$ .)

Step 4 — Bonus horizontal asymptote. Equal degrees $2 = 2$ , leading coefficients $1/1$ → HA $y = 1$ .

Verify: hole height $\frac{x + 1}{x + 3}$ at $x = 3$ gives $\frac{4}{6} = 0.6667$ ✓. Near $x = - 3$ : $x = - 2.99 \Rightarrow \frac{- 1.99}{0.01} = - 199$ (right side, $- \infty$ ) ✓; $x = - 3.01 \Rightarrow \frac{- 2.01}{- 0.01} = 201$ (left side, $+ \infty$ ) ✓. HA: $h (1000) = \frac{997997}{999991} \approx 0.998 \to 1$ ✓.

Ex 9 — Cell K: slant (oblique) asymptote

Worked example Find the slant asymptote of

$f (x) = \frac{x ^{2} + 3 x + 2}{x - 1}$ . Forecast: deg top $= 2$ , deg bottom $= 1$ , so deg top is exactly one more than deg bottom. There is no horizontal asymptote (top bigger → bust), but the curve does line up with a slanted straight line far out. Guess whether that line rises or falls.

Step 1 — Why polynomial division? Why this tool: when deg top $=$ deg bottom $+ 1$ , dividing the polynomials rewrites the fraction as $(line) + (small remainder)$ . The remainder is a proper fraction that $\to 0$ at infinity, so the line is exactly what the curve hugs. Division is the tool that peels the line off the top.

Step 2 — Do the long division of $x^{2} + 3 x + 2$ by $x - 1$ : $\frac{x ^{2} + 3 x + 2}{x - 1} = x + 4 + \frac{6}{x - 1} .$ (Check: $(x + 4) (x - 1) = x^{2} + 3 x - 4$ , plus remainder $6$ gives $x^{2} + 3 x + 2$ ✓.)

Step 3 — Send $x \to \pm \infty$ . Why: the term $\frac{6}{x - 1}$ is nonzero over a huge number → $0$ . What survives is the line $y = x + 4$ . So the slant asymptote is $y = x + 4$ on both ends (it rises, slope $+ 1$ ).

Step 4 — Note the vertical asymptote too. The denominator $x - 1 = 0$ at $x = 1$ with numerator $1 + 3 + 2 = 6 \neq = 0$ : nonzero ÷ zero → VA $x = 1$ . So this function has one slant and one vertical asymptote, no horizontal one.

Verify: at $x = 1000$ , $f (1000) = \frac{1003002}{999} \approx 1004.006$ , while $x + 4 = 1004$ — the gap is $\approx 0.006 \to 0$ ✓. At $x = - 1000$ , $f = \frac{997002}{- 1001} \approx - 996.006$ and $x + 4 = - 996$ , gap $\approx - 0.006 \to 0$ ✓. Remainder identity $(x + 4) (x - 1) + 6 = x^{2} + 3 x + 2$ ✓.

Recall Cover-the-matrix self-test

One-sided signs when denominator has an odd-power zero ::: opposite (limit DNE, still a VA) Even-power zero in denominator ::: same sign both sides Both top and bottom $\to 0$ ::: possible removable hole — factor first deg top $<$ deg bottom at infinity ::: HA $y = 0$ deg top $=$ deg bottom ::: HA = ratio of leading coefficients deg top $>$ deg bottom ::: no HA deg top $=$ deg bottom $+ 1$ ::: slant asymptote (found by polynomial division) $x^{2}$ as $x \to - \infty$ ::: $- x$ , flips the sign — two HAs possible

Connections

Infinite limits and limits at infinity — vertical - horizontal asymptotes
One-sided limits
Continuity and removable discontinuities
Slant (oblique) asymptotes via polynomial division
L'Hôpital's rule for indeterminate forms
Curve sketching using first and second derivatives
Limits — formal epsilon-delta definition

Worked examples — Infinite limits and limits at infinity — vertical - horizontal asymptotes

The scenario matrix

Ex 1 — Cell A: sides disagree