Connecting Infinite Limits and Vertical Asymptotes: Study Guide (Calculus AB 1.14)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so limits are written in a compact form:

• lim (x → a) f(x) means “the limit as x approaches a of f(x)”.
• x → a⁻ means x approaches a from the left (x < a). x → a⁺ means x approaches a from the right (x > a).
• ∞ is read “infinity”. It is a symbol for “without bound”, not a number you can calculate with.

On paper, write the usual form with “x → a⁺” under “lim”.

What an infinite limit means

Look at f(x) = 1/x² near x = 0.

x	±0.1	±0.01	±0.001
1/x²	100	10 000	1 000 000
1/x	±10	±100	±1000

As x gets closer to 0 from either side, 1/x² does not settle near any number. It becomes larger than any value you choose, and it stays positive. We record this as

lim (x → 0) 1/x² = ∞

This is an infinite limit. The limit idea from Topic 1.2 asked “which number do the values approach?” An infinite limit extends that idea: the values do not approach a number, but they do move in a definite way, growing without bound in one direction.

Be careful with what the statement claims. The limit does not exist as a real number, because ∞ is not a number. Writing “= ∞” is a precise way of saying how it fails to exist: the function is unbounded, and its values are positive near 0. “= −∞” means unbounded and negative.

Now look at 1/x. From the right, 1/x is positive and grows: lim (x → 0⁺) 1/x = ∞. From the left, 1/x is negative and grows in size: lim (x → 0⁻) 1/x = −∞. The two sides behave differently, so you cannot write a single two-sided statement such as “lim (x → 0) 1/x = ∞”. Give the two one-sided limits instead.

The difference comes from the power in the denominator. An odd power such as x or x³ changes sign at 0. An even power such as x² never goes negative.

Vertical asymptotes, defined with limits

Definition. The line x = a is a vertical asymptote of the graph of f if at least one of these is true: lim (x → a⁻) f(x) = ∞, lim (x → a⁻) f(x) = −∞, lim (x → a⁺) f(x) = ∞, lim (x → a⁺) f(x) = −∞.

So a vertical asymptote is not just “a place where the denominator is 0”. It is a claim about unbounded behaviour, and a one-sided infinite limit is the evidence. On a free-response answer, write that evidence down: “Because lim (x → 3⁺) f(x) = ∞, the line x = 3 is a vertical asymptote.”

Vertical asymptotes are not only for fractions:

Function	Infinite limit	Vertical asymptote
ln x	lim (x → 0⁺) ln x = −∞	x = 0 (graph exists only on the right)
tan x	lim (x → π/2⁻) tan x = ∞ and lim (x → π/2⁺) tan x = −∞	x = π/2 (and every odd multiple of π/2)
e^(1/x)	lim (x → 0⁺) e^(1/x) = ∞, but lim (x → 0⁻) e^(1/x) = 0	x = 0, from one side only

The last row shows why the definition says “at least one”. The left-hand limit is a finite number, yet x = 0 is still a vertical asymptote.

A vertical asymptote at x = a always means f is not continuous at a. This is the third type of discontinuity from Topic 1.10, and unlike a hole it cannot be removed by redefining one value (Topic 1.13): no single number can fill a gap that runs off to infinity.

Finding one-sided infinite limits: the sign check

When direct substitution at x = a gives a nonzero number over 0, the expression is unbounded near a. Topic 1.6 told you this means “no finite limit”. Now you can say more.

1. Substitute. Confirm the top tends to a nonzero number c and the bottom tends to 0.
2. Find the sign of the bottom on each side. Pick a value just to the left of a and just to the right, or reason from the factors.
3. Combine signs. (sign of c) × (sign of the bottom) gives the sign of f(x) on that side.
4. Write the limits. Positive and unbounded gives +∞; negative and unbounded gives −∞.

Example: (x + 4)/(x − 2) as x → 2. The top tends to 6, which is positive. For x slightly more than 2, x − 2 is a small positive number, so the fraction is large and positive: lim (x → 2⁺) = ∞. For x slightly less than 2, x − 2 is small and negative, so lim (x → 2⁻) = −∞.

Odd and even powers. If the factor (x − a) in the bottom appears to an even power, the bottom has the same sign on both sides, so both one-sided limits match. If it appears to an odd power, the sign flips and the one-sided limits are opposite.

Holes versus asymptotes

A zero of the denominator gives one of two pictures. Decide which by simplifying first.

Substitution at x = a gives	After cancelling common factors	Graph at x = a	Limit at a
0/0	The bottom is no longer 0 at a	Hole	A finite number (Topic 1.6)
0/0	The bottom is still 0 at a, top is not	Vertical asymptote	Infinite, check each side
nonzero/0	(nothing cancels)	Vertical asymptote	Infinite, check each side

For example, (x² − 9)/(x − 3) has a hole at x = 3 with limit 6, while (x + 3)/(x − 3) has a vertical asymptote at x = 3.

Worked example 1: a hole and an asymptote in one function

Question. Let f(x) = (x² + x − 6)/(x² − 5x + 6). Find every vertical asymptote of the graph of f and every hole. Describe the behaviour of f near each asymptote using limits.

1. Find where f is undefined. The bottom factors as (x − 2)(x − 3), so f is undefined at x = 2 and x = 3.
2. Factor the top. x² + x − 6 = (x − 2)(x + 3).
3. Cancel the common factor, valid for x ≠ 2: f(x) = (x + 3)/(x − 3) for x ≠ 2.
4. At x = 2. The simplified bottom is 2 − 3 = −1, not 0. So lim (x → 2) f(x) = (2 + 3)/(2 − 3) = −5 The limit is finite, so there is a hole at (2, −5), not an asymptote.
5. At x = 3. The simplified form gives 6/0: nonzero over 0.
   • Right of 3: x − 3 > 0 and x + 3 ≈ 6 > 0, so the fraction is positive. lim (x → 3⁺) f(x) = ∞.
   • Left of 3: x − 3 < 0, so the fraction is negative. lim (x → 3⁻) f(x) = −∞.
6. Conclude with the definition. Because lim (x → 3⁺) f(x) = ∞, the line x = 3 is a vertical asymptote. It is the only one.

Check. f(3.001) = 6.001/0.001 = 6001 and f(2.999) = 5.999/(−0.001) = −5999. Large, with the predicted signs.

Worked example 2: an even power gives a two-sided infinite limit

Question. Let g(x) = (2x − 7)/(x − 1)². Describe the behaviour of g as x → 1, and state whether x = 1 is a vertical asymptote.

1. Substitute. Top: 2(1) − 7 = −5. Bottom: (1 − 1)² = 0. Nonzero over 0, so g is unbounded near 1.
2. Sign of the bottom. (x − 1)² is a square, so it is positive on both sides of 1 (for x ≠ 1).
3. Combine signs. Negative top ÷ small positive bottom = large negative number, on both sides.
4. Write the limits. lim (x → 1⁻) g(x) = −∞ and lim (x → 1⁺) g(x) = −∞ The two sides agree, so you may write the two-sided statement lim (x → 1) g(x) = −∞.
5. Conclude. Since lim (x → 1) g(x) = −∞, the line x = 1 is a vertical asymptote. Near x = 1 the graph plunges downward on both sides.

Check. g(1.01) = −4.98/0.0001 = −49 800 and g(0.99) = −5.02/0.0001 = −50 200. Both large and negative.

Interpretation. “lim (x → 1) g(x) = −∞” still means the limit does not exist as a number. The statement just tells you the way it fails: every value near 1 is a large negative number.

Worked example 3: a vertical asymptote without a polynomial

Question. Let q(x) = x/ln x, for x > 0 and x ≠ 1. Show that x = 1 is a vertical asymptote of the graph of q, and decide whether x = 0 is one too.

1. Where is q undefined? ln x is defined only for x > 0, and ln x = 0 only at x = 1. So the candidates are x = 1 and the edge x = 0.
2. At x = 1, substitute. Top: 1. Bottom: ln 1 = 0. Nonzero over 0.
3. Sign of ln x near 1. ln x > 0 for x > 1, and ln x < 0 for 0 < x < 1.
4. Write the limits.
   • lim (x → 1⁺) x/ln x = ∞ (positive ÷ small positive).
   • lim (x → 1⁻) x/ln x = −∞ (positive ÷ small negative).
5. Justify. Because lim (x → 1⁺) q(x) = ∞, the line x = 1 is a vertical asymptote of the graph of q.
6. At x = 0 (from the right only). The top x → 0. The bottom ln x → −∞, so its size grows without bound. A quantity near 0 divided by a huge number is near 0: lim (x → 0⁺) x/ln x = 0. That is finite, so x = 0 is not a vertical asymptote, even though q is undefined there.

Check. q(1.01) ≈ 101.5 and q(0.99) ≈ −98.5. The signs and sizes fit.

Step 6 is the lesson: being undefined at a point is not enough. You need an infinite limit.

Common misconceptions

• “lim (x → a) f(x) = ∞ means the limit exists and equals infinity.” The limit does not exist. “= ∞” names the way it fails: unbounded, positive.
• “Every zero of the denominator is a vertical asymptote.” Not if the factor cancels. (x² − 9)/(x − 3) has a hole at x = 3. Simplify first.
• “Nonzero over 0 means the limit is 0.” A fixed number divided by something tiny is huge, not 0.
• Writing a two-sided infinite limit when the sides disagree. For 1/x at 0, the sides go to +∞ and −∞. Write the one-sided limits separately.
• Guessing the sign. Do not assume +∞. Check the sign of the top and of the bottom on each side. A squared factor in the bottom is positive on both sides.
• “A vertical asymptote needs both sides to be infinite.” One infinite one-sided limit is enough, as with ln x at 0 or e^(1/x) at 0.
• “The function can never be defined at an asymptote.” The definition is about nearby values. A function such as f(x) = 1/x² for x ≠ 0, with f(0) = 5, is defined at 0, yet x = 0 is still a vertical asymptote because lim (x → 0) f(x) = ∞.
• Trusting a calculator graph alone. A graphing window can draw a near-vertical line joining the two branches, or miss the asymptote between plotted points. Use the sign check to confirm.

Where this leads

Topic 1.15, Connecting Limits at Infinity and Horizontal Asymptotes, turns the question around: instead of x → a with f(x) unbounded, it asks what f(x) does as x itself grows without bound. Infinite limits return later too. In Unit 5 they help you sketch graphs, and in Unit 6 improper integrals (BC only) handle areas next to a vertical asymptote. The previous topic, Removing Discontinuities, explains why a hole can be filled but an asymptote cannot. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.